数学分析积分表及常用积分公式

一.含有xnx^nxn的形式

1.∫xndx=xn+1n+1+C(n≠−1)1.\int x^ndx=\frac{x^{n+1}}{n+1}+C\,(n≠-1)1.∫xndx=n+1xn+1+C(n=−1)

2.∫dxx=ln∣x∣+C2.\int\frac{dx}{x}=ln\,|x|+C2.∫xdx=ln∣x∣+C

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二.含有ax+bax+bax+b的形式

1.∫dxax+b=1aln⁡∣ax+b∣+C1.\int{\frac{dx}{ax+b}}=\frac{1}{a}\ln{\mid ax+b \mid}+C1.∫ax+bdx=a1ln∣ax+b∣+C

2.∫(ax+b)ndx=1a(n+1)(ax+b)n+1+C(n≠−1)2.\int{(ax+b)^ndx}=\frac{1}{a(n+1)}(ax+b)^{n+1}+C(n\neq-1)2.∫(ax+b)ndx=a(n+1)1(ax+b)n+1+C(n=−1)

3.∫xax+bdx=1a2(ax−bln⁡∣ax+b∣)+C3.\int{\frac{x}{ax+b}dx}=\frac{1}{a^2}(ax-b\ln{\mid ax+b \mid})+C3.∫ax+bxdx=a21(ax−bln∣ax+b∣)+C

4.∫x2ax+bdx=1a3[12(ax+b)2−2b(ax+b)+b2ln⁡∣ax+b∣]+C4.\int{\frac{x^2}{ax+b}dx}=\frac{1}{a^3}[\frac{1}{2}(ax+b)^2-2b(ax+b)+b^2\ln{\mid ax+b \mid}]+C4.∫ax+bx2dx=a31[21(ax+b)2−2b(ax+b)+b2ln∣ax+b∣]+C

5.∫dxx(ax+b)=−1bln⁡∣ax+bx∣+C5.\int{\frac{dx}{x(ax+b)}}=-\frac{1}{b}\ln{\mid \frac{ax+b}{x} \mid}+C5.∫x(ax+b)dx=−b1ln∣xax+b∣+C

6.∫dxx2(ax+b)=−1bx+ab2ln⁡∣ax+bx∣+C6.\int{\frac{dx}{x^2(ax+b)}}=-\frac{1}{bx}+\frac{a}{b^2}\ln{\mid \frac{ax+b}{x} \mid}+C6.∫x2(ax+b)dx=−bx1+b2aln∣xax+b∣+C

7.∫x(ax+b)2dx=1a2(ln⁡∣ax+b∣+bax+b)+C7.\int{\frac{x}{(ax+b)^2}dx}=\frac{1}{a^2}(\ln{\mid ax+b \mid}+\frac{b}{ax+b})+C7.∫(ax+b)2xdx=a21(ln∣ax+b∣+ax+bb)+C

8.∫x2(ax+b)2dx=1a3(ax+b−2bln⁡∣ax+b∣−b2ax+b)+C8.\int{\frac{x^2}{(ax+b)^2}dx}=\frac{1}{a^3}(ax+b-2b\ln{\mid ax+b \mid-\frac{b^2}{ax+b}})+C8.∫(ax+b)2x2dx=a31(ax+b−2bln∣ax+b∣−ax+bb2)+C

9.∫dxx(ax+b)2=1b(ax+b)−1b2ln⁡∣ax+bx∣+C9.\int{\frac{dx}{x(ax+b)^2}}=\frac{1}{b(ax+b)}-\frac{1}{b^2}\ln{\mid \frac{ax+b}{x} \mid}+C9.∫x(ax+b)2dx=b(ax+b)1−b21ln∣xax+b∣+C

10.∫dxx2(ax+b)2=−1b2[2ax+bx(ax+b)+2abln⁡∣xax+b∣]+C10.\int\frac{dx}{x^2(ax+b)^2}=-\frac{1}{b^2}[\frac{2ax+b}{x(ax+b)}+\frac{2a}{b}\ln|\frac{x}{ax+b}|]+C10.∫x2(ax+b)2dx=−b21[x(ax+b)2ax+b+b2aln∣ax+bx∣]+C

11.∫x2(ax+b)3dx=1a3[2bax+b−b22(ax+b)2+ln⁡∣ax+b∣]+C11.\int\frac{x^2}{(ax+b)^3}dx=\frac{1}{a^3}[\frac{2b}{ax+b}-\frac{b^2}{2(ax+b)^2}+\ln|ax+b|]+C11.∫(ax+b)3x2dx=a31[ax+b2b−2(ax+b)2b2+ln∣ax+b∣]+C

12.∫x(ax+b)ndx=1a2[−1(n−2)(ax+b)n−2+b(n−1)(ax+b)n−1]+C(n≠1,2)12.\int\frac{x}{(ax+b)^n}dx=\frac{1}{a^2}[-\frac{1}{(n-2)(ax+b)^{n-2}}+\frac{b}{(n-1)(ax+b)^{n-1}}]+C\,(n≠1,2)12.∫(ax+b)nxdx=a21[−(n−2)(ax+b)n−21+(n−1)(ax+b)n−1b]+C(n=1,2)

13.∫x2(ax+b)ndx=1a3[−1(n−3)(ax+b)n−3+2b(n−2)(ax+b)n−2−b2(n−1)(ax+b)n−1]+C(n≠1,2,3)13.\int\frac{x^2}{(ax+b)^n}dx=\frac{1}{a^3}[-\frac{1}{(n-3)(ax+b)^{n-3}}+\frac{2b}{(n-2)(ax+b)^{n-2}}-\frac{b^2}{(n-1)(ax+b)^{n-1}}]+C\,(n≠1,2,3)13.∫(ax+b)nx2dx=a31[−(n−3)(ax+b)n−31+(n−2)(ax+b)n−22b−(n−1)(ax+b)n−1b2]+C(n=1,2,3)

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三.含有ax2+b(a>0)ax^2+b\,(a>0)ax2+b(a>0)的形式

1.∫dxax2+bdx={1abarctan⁡abx+C(b>0)12−abln⁡∣x−ax+a∣+C(b<0)1.\int{\frac{dx}{ax^2+b}dx}=\begin{cases}\frac{1}{\sqrt{ab}}\arctan{\sqrt{\frac{a}{b}}x}+C(b>0)\\\frac{1}{2{\sqrt{-ab}}}\ln{\mid\frac{x-a}{x+a} \mid}+C(b<0) \end{cases}1.∫ax2+bdxdx={ab1arctanbax+C(b>0)2−ab1ln∣x+ax−a∣+C(b<0)

2.∫xax2+bdx=12aln⁡∣ax2+b∣+C2.\int\frac{x}{ax^2+b}dx=\frac{1}{2a}\ln|ax^2+b|+C2.∫ax2+bxdx=2a1ln∣ax2+b∣+C

3.∫x2ax2+bdx=xa−ba∫dxax2+b3.\int\frac{x^2}{ax^2+b}dx=\frac{x}{a}-\frac{b}{a}\int\frac{dx}{ax^2+b}3.∫ax2+bx2dx=ax−ab∫ax2+bdx

4.∫dxx(ax2+b)=12bln⁡x2∣ax2+b∣+C4.\int\frac{dx}{x(ax^2+b)}=\frac{1}{2b}\ln\frac{x^2}{|ax^2+b|}+C4.∫x(ax2+b)dx=2b1ln∣ax2+b∣x2+C

5.∫dxx2(ax2+b)=−1bx−ab∫dxax2+b5.\int\frac{dx}{x^2(ax^2+b)}=-\frac{1}{bx}-\frac{a}{b}\int\frac{dx}{ax^2+b}5.∫x2(ax2+b)dx=−bx1−ba∫ax2+bdx

6.∫dxx3(ax2+b)=a2b2ln⁡∣ax2+b∣x2−12bx2+C6.\int\frac{dx}{x^3(ax^2+b)}=\frac{a}{2b^2}\ln\frac{|ax^2+b|}{x^2}-\frac{1}{2bx^2}+C6.∫x3(ax2+b)dx=2b2alnx2∣ax2+b∣−2bx21+C

7.∫dx(ax2+b)2=x2b(ax2+b)+12b∫dxax2+b7.\int\frac{dx}{(ax^2+b)^2}=\frac{x}{2b(ax^2+b)}+\frac{1}{2b}\int\frac{dx}{ax^2+b}7.∫(ax2+b)2dx=2b(ax2+b)x+2b1∫ax2+bdx

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四.含有±x2±a2(a>0)±x^2±a^2\,(a>0)±x2±a2(a>0)的形式
1.∫dxx2+a2=1aarctan⁡xa+C1.\int{\frac{dx}{x^2+a^2}}=\frac{1}{a}\arctan{\frac{x}{a}}+C1.∫x2+a2dx=a1arctanax+C

2.∫dx(a2±x2)n=12a2(n−1)[x(a2±x2)n−1+(2n−3)∫dx(a2±x2)n−1](n≠1)2.\int{\frac{dx}{(a^2±x^2)^n}}=\frac{1}{2a^2(n-1)}[\frac{x}{(a^2±x^2)^{n-1}}+(2n-3)\int{\frac{dx}{(a^2±x^2)^{n-1}}}]\,(n≠1)2.∫(a2±x2)ndx=2a2(n−1)1[(a2±x2)n−1x+(2n−3)∫(a2±x2)n−1dx](n=1)

3.∫dxx2−a2=12aln⁡∣x−ax+a∣+C3.\int{\frac{dx}{x^2-a^2}}=\frac{1}{2a}\ln{\mid \frac{x-a}{x+a} \mid}+C3.∫x2−a2dx=2a1ln∣x+ax−a∣+C

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五.含有ax2+bx+cax^2+bx+cax2+bx+c的形式
1.∫dxax2+bx+c={24ac−b2arctan2ax+b4ac−b2+C(b2<4ac)1b2−4acln∣2ax+b−b2−4ac2ax+b+b2−4ac∣+C(b2>4ac)1.\int \frac{dx}{ax^2+bx+c}=\begin{cases}\frac{2}{\sqrt{4ac-b^2}}arctan\frac{2ax+b}{\sqrt{4ac-b^2}}+C\,(b^2<4ac)\\ \frac{1}{b^2-4ac}ln|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}|+C\,(b^2>4ac)\end{cases}1.∫ax2+bx+cdx={4ac−b22arctan4ac−b22ax+b+C(b2<4ac)b2−4ac1ln∣2ax+b+b2−4ac2ax+b−b2−4ac∣+C(b2>4ac)

2.∫xax2+bx+cdx=12aln∣ax2+bx+c∣−b2a∫dxax2+bx+c2.\int \frac{x}{ax^2+bx+c}dx=\frac{1}{2a}ln|ax^2+bx+c|-\frac{b}{2a}\int \frac{dx}{ax^2+bx+c}2.∫ax2+bx+cxdx=2a1ln∣ax2+bx+c∣−2ab∫ax2+bx+cdx

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六.含有ax+b\sqrt{ax+b}ax+b的形式

1.∫ax+bdx1.\int{\sqrt{ax+b}dx}1.∫ax+bdx = 23a(ax+b)3+C\frac{2}{3a}\sqrt{(ax+b)^3}+C3a2(ax+b)3+C

2.∫xax+bdx2.\int{x\sqrt{ax+b}dx}2.∫xax+bdx = 215a2(3ax−2b)(ax+b)3+C\frac{2}{15a^2}(3ax-2b)\sqrt{(ax+b)^3}+C15a22(3ax−2b)(ax+b)3+C

3.∫x2ax+bdx3.\int{x^2\sqrt{ax+b}dx}3.∫x2ax+bdx = 215a2(15a2x2−12abx+8b2)(ax+b)3+C\frac{2}{15a^2}(15a^2x^2-12abx+8b^2)\sqrt{(ax+b)^3}+C15a22(15a2x2−12abx+8b2)(ax+b)3+C

4.∫xnax+bdx=2a(2n+3)[xn(ax+b)32−nb∫xn−1ax+bdx]4.\int{x^n\sqrt{ax+b}}dx=\frac{2}{a(2n+3)}[x^n(ax+b)^{\frac{3}{2}}-nb\int x^{n-1}\sqrt{ax+b}dx]4.∫xnax+bdx=a(2n+3)2[xn(ax+b)23−nb∫xn−1ax+bdx]

5.∫xax+bdx=23a2(ax−2b)ax+b+C5.\int{\frac{x}{\sqrt{ax+b}}dx}=\frac{2}{3a^2}(ax-2b)\sqrt{ax+b}+C5.∫ax+bxdx=3a22(ax−2b)ax+b+C

6.∫x2ax+bdx=215a3(3a2x2−4abcx+8b2)ax+b+C6.\int{\frac{x^2}{\sqrt{ax+b}}dx}=\frac{2}{15a^3}(3a^2x^2-4abcx+8b^2)\sqrt{ax+b}+C6.∫ax+bx2dx=15a32(3a2x2−4abcx+8b2)ax+b+C

7.∫xnax+bdx=2(2n+1)a(xnax+b−nb∫xn−1ax+bdx)7.\int\frac{x^n}{\sqrt{ax+b}}dx=\frac{2}{(2n+1)a}(x^n\sqrt{ax+b}-nb\int\frac{x^{n-1}}{\sqrt{ax+b}}dx)7.∫ax+bxndx=(2n+1)a2(xnax+b−nb∫ax+bxn−1dx)

8.∫dxxax+b={1bln⁡∣ax+b−bax+b+b∣+C(b>0)2−barctan⁡ax+b−b+C(b<0)8.\int{\frac{dx}{x\sqrt{ax+b}}}=\begin{cases} \frac{1}{\sqrt{b}}\ln{\mid \frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}} \mid}+C(b>0)\\\frac{2}{\sqrt{-b}}\arctan{\sqrt{\frac{ax+b}{-b}}}+C(b<0) \end{cases}8.∫xax+bdx=⎩⎨⎧b1ln∣ax+b+bax+b−b∣+C(b>0)−b2arctan−bax+b+C(b<0)

9.∫dxx2ax+bdx=−ax+bbx−a2b∫dxxax+b9.\int{\frac{dx}{x^2\sqrt{ax+b}}dx}=-\frac{\sqrt{ax+b}}{bx}-\frac{a}{2b}\int{\frac{dx}{x\sqrt{ax+b}}}9.∫x2ax+bdxdx=−bxax+b−2ba∫xax+bdx

10.∫dxxnax+b=−1b(n−1)[ax+bxn−1+a(2n−3)2∫dxxn−1ax+b](n≠−1)10.\int\frac{dx}{x^n\sqrt{ax+b}}=-\frac{1}{b(n-1)}[\frac{\sqrt{ax+b}}{x^{n-1}}+\frac{a(2n-3)}{2}\int\frac{dx}{x^{n-1}\sqrt{ax+b}}]\,(n≠-1)10.∫xnax+bdx=−b(n−1)1[xn−1ax+b+2a(2n−3)∫xn−1ax+bdx](n=−1)

11.∫ax+bxdx=2ax+b+b∫dxxax+b11.\int{\frac{\sqrt{ax+b}}{x}dx}=2\sqrt{ax+b}+b\int{\frac{dx}{x\sqrt{ax+b}}}11.∫xax+bdx=2ax+b+b∫xax+bdx

12.∫ax+bx2dx=−ax+bx+a2∫dxxax+b12.\int{\frac{\sqrt{ax+b}}{x^2}dx}=-\frac{\sqrt{ax+b}}{x}+\frac{a}{2}\int{\frac{dx}{x\sqrt{ax+b}}}12.∫x2ax+bdx=−xax+b+2a∫xax+bdx

13.∫ax+bxndx=−1b(n−1)[(ax+b)32xn−1+(2n−5)a2∫ax+bxn−1dx](n≠−1)13.\int{\frac{\sqrt{ax+b}}{x^n}dx}=-\frac{1}{b(n-1)}[\frac{(ax+b)^{\frac{3}{2}}}{x^{n-1}}+\frac{(2n-5)a}{2}\int{\frac{\sqrt{ax+b}}{x^{n-1}}dx}]\,(n≠-1)13.∫xnax+bdx=−b(n−1)1[xn−1(ax+b)23+2(2n−5)a∫xn−1ax+bdx](n=−1)

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七.含有±ax2+bx+c(a>0)\sqrt{±ax^2+bx+c}\,(a>0)±ax2+bx+c(a>0)的形式

1.∫dxax2+bx+c=1aln⁡∣2ax+b+2aax2+bx+c∣+C1.\int\frac{dx}{\sqrt{ax^2+bx+c}}=\frac{1}{\sqrt{a}}\ln|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}|+C1.∫ax2+bx+cdx=a1ln∣2ax+b+2aax2+bx+c∣+C

2.∫ax2+bx+cdx=2ax+b4aax2+bx+c+4ac−b28a32ln⁡∣2ax+b+2aax2+bx+c∣+C2.\int\sqrt{ax^2+bx+c}dx=\frac{2ax+b}{4a}\sqrt{ax^2+bx+c}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}|+C2.∫ax2+bx+cdx=4a2ax+bax2+bx+c+8a234ac−b2ln∣2ax+b+2aax2+bx+c∣+C

3.∫xax2+bx+cdx=1aax2+bx+c−b2a32ln⁡∣2ax+b+2aax2+bx+c∣+C3.\int\frac{x}{\sqrt{ax^2+bx+c}}dx=\frac{1}{a}\sqrt{ax^2+bx+c}-\frac{b}{2a^{\frac{3}{2}}}\ln|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}|+C3.∫ax2+bx+cxdx=a1ax2+bx+c−2a23bln∣2ax+b+2aax2+bx+c∣+C

4.∫dx−ax2+bx+c=1aarcsin⁡2ax−bb2+4ac+C4.\int\frac{dx}{\sqrt{-ax^2+bx+c}}=\frac{1}{\sqrt{a}}\arcsin\frac{2ax-b}{\sqrt{b^2+4ac}}+C4.∫−ax2+bx+cdx=a1arcsinb2+4ac2ax−b+C

5.∫−ax2+bx+cdx=2ax−b4a−ax2+bx+c+4ac+b28a32arcsin⁡2ax−bb2+4ac+C5.\int\sqrt{-ax^2+bx+c}dx=\frac{2ax-b}{4a}\sqrt{-ax^2+bx+c}+\frac{4ac+b^2}{8a^{\frac{3}{2}}}\arcsin\frac{2ax-b}{\sqrt{b^2+4ac}}+C5.∫−ax2+bx+cdx=4a2ax−b−ax2+bx+c+8a234ac+b2arcsinb2+4ac2ax−b+C

6.∫x−ax2+bx+cdx=−1a−ax2+bx+c+b2a32arcsin⁡2ax−bb2+4ac+C6.\int\frac{x}{\sqrt{-ax^2+bx+c}}dx=-\frac{1}{a}\sqrt{-ax^2+bx+c}+\frac{b}{2a^{\frac{3}{2}}}\arcsin\frac{2ax-b}{\sqrt{b^2+4ac}}+C6.∫−ax2+bx+cxdx=−a1−ax2+bx+c+2a23barcsinb2+4ac2ax−b+C

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八.含有x2±a2(a>0)\sqrt{x^2±a^2}(a>0)x2±a2(a>0)的形式

1.∫x2±a2dx=12(xx2±a2±a2ln∣x+x2±a2∣)+C1.\int\sqrt{x^2±a^2}dx=\frac{1}{2}(x\sqrt{x^2±a^2}±a^2ln\,|x+\sqrt{x^2±a^2}|)+C1.∫x2±a2dx=21(xx2±a2±a2ln∣x+x2±a2∣)+C

2.∫(x2+a2)32=x8(2x2+5a2)x2+a2+38a4ln⁡(x+x2+a2)+C2.\int(x^2+a^2)^{\frac{3}{2}}=\frac{x}{8}(2x^2+5a^2)\sqrt{x^2+a^2}+\frac{3}{8}a^4\ln(x+\sqrt{x^2+a^2})+C2.∫(x2+a2)23=8x(2x2+5a2)x2+a2+83a4ln(x+x2+a2)+C

3.∫(x2−a2)32=x8(2x2−5a2)x2−a2+38a4ln⁡∣x+x2−a2∣+C3.\int(x^2-a^2)^{\frac{3}{2}}=\frac{x}{8}(2x^2-5a^2)\sqrt{x^2-a^2}+\frac{3}{8}a^4\ln|x+\sqrt{x^2-a^2}|+C3.∫(x2−a2)23=8x(2x2−5a2)x2−a2+83a4ln∣x+x2−a2∣+C

4.∫xx2±a2dx=13(x2±a2)32+C4.\int x\sqrt{x^2±a^2}dx=\frac{1}{3}(x^2±a^2)^{\frac{3}{2}}+C4.∫xx2±a2dx=31(x2±a2)23+C

5.∫x2x2±a2dx=18[x(2x2±a2)x2±a2−a4ln∣x+x2±a2∣]+C5.\int x^2\sqrt{x^2±a^2}dx=\frac{1}{8}[x(2x^2±a^2)\sqrt{x^2±a^2}-a^4ln\,|x+\sqrt{x^2±a^2}|]+C5.∫x2x2±a2dx=81[x(2x2±a2)x2±a2−a4ln∣x+x2±a2∣]+C

6.∫1xx2+a2dx=x2+a2−aln∣a+x2+a2x∣+C6.\int\frac{1}{x}\sqrt{x^2+a^2}dx=\sqrt{x^2+a^2}-aln\,|\frac{a+\sqrt{x^2+a^2}}{x}|+C6.∫x1x2+a2dx=x2+a2−aln∣xa+x2+a2∣+C

7.∫1xx2−a2dx=x2−a2−aarccos⁡ax+C7.\int\frac{1}{x}\sqrt{x^2-a^2}dx=\sqrt{x^2-a^2}-a\arccos\frac{a}{x}+C7.∫x1x2−a2dx=x2−a2−aarccosxa+C

8.∫1x2x2±a2dx=−1xx2±a2+ln∣x+x2±a2∣+C8.\int\frac{1}{x^2}\sqrt{x^2±a^2}dx=-\frac{1}{x}\sqrt{x^2±a^2}+ln\,|x+\sqrt{x^2±a^2}|+C8.∫x21x2±a2dx=−x1x2±a2+ln∣x+x2±a2∣+C

9.∫dxx2+a2=arshxa+C=ln∣x+x2+a2∣+C9.\int\frac{dx}{\sqrt{x^2+a^2}}=arsh\frac{x}{a}+C=ln\,|x+\sqrt{x^2+a^2}|+C9.∫x2+a2dx=arshax+C=ln∣x+x2+a2∣+C

10.∫dxx2−a2=x∣x∣arsh∣x∣a+C=ln∣x+x2−a2∣+C10.\int\frac{dx}{\sqrt{x^2-a^2}}=\frac{x}{|x|}arsh\frac{|x|}{a}+C=ln\,|x+\sqrt{x^2-a^2}|+C10.∫x2−a2dx=∣x∣xarsha∣x∣+C=ln∣x+x2−a2∣+C

11.∫xx2±a2dx=x2±a2+C11.\int\frac{x}{\sqrt{x^2±a^2}}dx=\sqrt{x^2±a^2}+C11.∫x2±a2xdx=x2±a2+C

12.∫x2x2±a2dx=12(xx2±a2∓a2ln∣x+x2±a2∣)+C12.\int\frac{x^2}{\sqrt{x^2±a^2}}dx=\frac{1}{2}(x\sqrt{x^2±a^2}∓a^2ln\,|x+\sqrt{x^2±a^2}|)+C12.∫x2±a2x2dx=21(xx2±a2∓a2ln∣x+x2±a2∣)+C

13.∫dxxx2+a2=−1aln∣a+x2+a2x∣+C13.\int\frac{dx}{x\sqrt{x^2+a^2}}=-\frac{1}{a}ln\,|\frac{a+\sqrt{x^2+a^2}}{x}|+C13.∫xx2+a2dx=−a1ln∣xa+x2+a2∣+C

14.∫dxxx2−a2=1aarccos⁡ax+C14.\int\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\arccos\frac{a}{x}+C14.∫xx2−a2dx=a1arccosxa+C

15.∫dxx2x2±a2=∓x2±a2a2x+C15.\int\frac{dx}{x^2\sqrt{x^2±a^2}}=∓\frac{\sqrt{x^2±a^2}}{a^2x}+C15.∫x2x2±a2dx=∓a2xx2±a2+C

16.∫dx(x2±a2)32=±xa2x2±a2+C16.\int\frac{dx}{(x^2±a^2)^{\frac{3}{2}}}=\frac{±x}{a^2\sqrt{x^2±a^2}}+C16.∫(x2±a2)23dx=a2x2±a2±x+C

17.∫x(x2±a2)32dx=−1x2±a2+C17.\int\frac{x}{(x^2±a^2)^{\frac{3}{2}}}dx=-\frac{1}{\sqrt{x^2±a^2}}+C17.∫(x2±a2)23xdx=−x2±a21+C

18.∫x2(x2±a2)32dx=−xx2±a2+ln⁡(x+x2±a2)+C18.\int\frac{x^2}{(x^2±a^2)^{\frac{3}{2}}}dx=-\frac{x}{\sqrt{x^2±a^2}}+\ln(x+\sqrt{x^2±a^2})+C18.∫(x2±a2)23x2dx=−x2±a2x+ln(x+x2±a2)+C

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九.含有a2−x2(a>0)\sqrt{a^2-x^2}(a>0)a2−x2(a>0)的形式*

1.∫a2−x2dx=12(xa2−x2+a2arcsin⁡xa)+C1.\int\sqrt{a^2-x^2}dx=\frac{1}{2}(x\sqrt{a^2-x^2}+a^2\arcsin\frac{x}{a})+C1.∫a2−x2dx=21(xa2−x2+a2arcsinax)+C

2.∫(a2−x2)32dx=58(5a2−2x2)a2−x2+38a4arcsin⁡xa+C2.\int(a^2-x^2)^{\frac{3}{2}}dx=\frac{5}{8}(5a^2-2x^2)\sqrt{a^2-x^2}+\frac{3}{8}a^4\arcsin\frac{x}{a}+C2.∫(a2−x2)23dx=85(5a2−2x2)a2−x2+83a4arcsinax+C

3.∫xa2−x2dx=−13(a2−x2)32+C3.\int x\sqrt{a^2-x^2}dx=-\frac{1}{3}(a^2-x^2)^{\frac{3}{2}}+C3.∫xa2−x2dx=−31(a2−x2)23+C

4.∫x2a2−x2dx=18[x(2x2−a2)a2−x2+a4arcsin⁡xa]+C4.\int x^2\sqrt{a^2-x^2}dx=\frac{1}{8}[x(2x^2-a^2)\sqrt{a^2-x^2}+a^4\arcsin\frac{x}{a}]+C4.∫x2a2−x2dx=81[x(2x2−a2)a2−x2+a4arcsinax]+C

5.∫1xa2−x2dx=a2−x2−aln⁡∣a+a2−x2x∣+C5.\int\frac{1}{x}\sqrt{a^2-x^2}dx=\sqrt{a^2-x^2}-a\ln|\frac{a+\sqrt{a^2-x^2}}{x}|+C5.∫x1a2−x2dx=a2−x2−aln∣xa+a2−x2∣+C

6.∫1x2a2−x2dx=−1xa2−x2−arcsin⁡xa+C6.\int\frac{1}{x^2}\sqrt{a^2-x^2}dx=-\frac{1}{x}\sqrt{a^2-x^2}-\arcsin\frac{x}{a}+C6.∫x21a2−x2dx=−x1a2−x2−arcsinax+C

7.∫dxa2−x2arcsin⁡xa+C7.\int\frac{dx}{\sqrt{a^2-x^2}}\arcsin\frac{x}{a}+C7.∫a2−x2dxarcsinax+C

8.∫dxxa2−x2=−1aln⁡∣a+a2−x2x∣+C8.\int\frac{dx}{x\sqrt{a^2-x^2}}=-\frac{1}{a}\ln|\frac{a+\sqrt{a^2-x^2}}{x}|+C8.∫xa2−x2dx=−a1ln∣xa+a2−x2∣+C

9.∫dxx2a2−x2=−a2−x2a2x+C9.\int\frac{dx}{x^2\sqrt{a^2-x^2}}=-\frac{\sqrt{a^2-x^2}}{a^2x}+C9.∫x2a2−x2dx=−a2xa2−x2+C

10.∫xa2−x2dx=−a2−x2+C10.\int\frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}+C10.∫a2−x2xdx=−a2−x2+C

11.∫x2a2−x2dx=12(−xa2−x2+a2arcsin⁡xa)+C11.\int\frac{x^2}{\sqrt{a^2-x^2}}dx=\frac{1}{2}(-x\sqrt{a^2-x^2}+a^2\arcsin\frac{x}{a})+C11.∫a2−x2x2dx=21(−xa2−x2+a2arcsinax)+C

12.∫dx(a2−x2)32=xa2a2−x2+C12.\int\frac{dx}{(a^2-x^2)^{\frac{3}{2}}}=\frac{x}{a^2\sqrt{a^2-x^2}}+C12.∫(a2−x2)23dx=a2a2−x2x+C

13.∫x(a2−x2)32dx=1a2−x2+C13.\int\frac{x}{(a^2-x^2)^{\frac{3}{2}}}dx=\frac{1}{\sqrt{a^2-x^2}}+C13.∫(a2−x2)23xdx=a2−x21+C

14.∫x2(a2−x2)32dx=xa2−x2−arcsin⁡xa+C14.\int\frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}dx=\frac{x}{\sqrt{a^2-x^2}}-\arcsin\frac{x}{a}+C14.∫(a2−x2)23x2dx=a2−x2x−arcsinax+C

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十.含有±x−ax−b\sqrt{±\frac{x-a}{x-b}}±x−bx−a或(x−a)(b−x)\sqrt{(x-a)(b-x)}(x−a)(b−x)的形式

1.∫x−ax−b=(x−b)x−ax−b+(b−a)ln⁡(∣x−a∣+∣x−b∣)+C1.\int\sqrt{\frac{x-a}{x-b}}=(x-b)\sqrt{\frac{x-a}{x-b}}+(b-a)\ln(\sqrt{|x-a|}+\sqrt{|x-b|})+C1.∫x−bx−a=(x−b)x−bx−a+(b−a)ln(∣x−a∣+∣x−b∣)+C

2.∫−x−ax−b=(x−b)−x−ax−b+(b−a)arcsin⁡x−ab−a+C2.\int\sqrt{-\frac{x-a}{x-b}}=(x-b)\sqrt{-\frac{x-a}{x-b}}+(b-a)\arcsin\sqrt{\frac{x-a}{b-a}}+C2.∫−x−bx−a=(x−b)−x−bx−a+(b−a)arcsinb−ax−a+C

3.∫dx(x−a)(b−x)=2arcsin⁡x−ab−a+C(a<b)3.\int\frac{dx}{\sqrt{(x-a)(b-x)}}=2\arcsin\sqrt{\frac{x-a}{b-a}}+C\,(a<b)3.∫(x−a)(b−x)dx=2arcsinb−ax−a+C(a<b)

4.∫(x−a)(b−x)dx=2x−a−b4(x−a)(b−x)+(b−a)24arcsin⁡x−ab−a+C(a<b)4.\int\sqrt{(x-a)(b-x)}dx=\frac{2x-a-b}{4}\sqrt{(x-a)(b-x)}+\frac{(b-a)^2}{4}\arcsin\sqrt{\frac{x-a}{b-a}}+C\,(a<b)4.∫(x−a)(b−x)dx=42x−a−b(x−a)(b−x)+4(b−a)2arcsinb−ax−a+C(a<b)

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十一.含有sinx,cosxsin\,x,cos\,xsinx,cosx的形式

1.∫sinxdx=−cosx+C1.\int sin\,x\,dx=-cos\,x+C1.∫sinxdx=−cosx+C

2.∫cosxdx=sinx+C2.\int cos\,x\,dx=sin\,x+C2.∫cosxdx=sinx+C

3.∫sin2xdx=x2−14sin2x+C3.\int sin^2x\,dx=\frac{x}{2}-\frac{1}{4}sin\,2x+C3.∫sin2xdx=2x−41sin2x+C

4.∫cos2xdx=x2+14sin2x+C4.\int cos^2x\,dx=\frac{x}{2}+\frac{1}{4}sin\,2x+C4.∫cos2xdx=2x+41sin2x+C

5.∫sinnxdx=−1nsinn−1xcosx+n−1n∫sinn−2xdx5.\int sin^nx\,dx=-\frac{1}{n}sin^{n-1}x\,cos\,x+\frac{n-1}{n}\int sin^{n-2}x\,dx5.∫sinnxdx=−n1sinn−1xcosx+nn−1∫sinn−2xdx

6.∫cosnxdx=1ncosn−1xsinx+n−1n∫cosn−2xdx6.\int cos^nx\,dx=\frac{1}{n}cos^{n-1}x\,sin\,x+\frac{n-1}{n}\int cos^{n-2}x\,dx6.∫cosnxdx=n1cosn−1xsinx+nn−1∫cosn−2xdx

7.∫cosmxsinnxdx=1m+ncosm−1xsinn+1x+m−1m+n∫cosm−2sinnxdx=−1m+ncosm+1xsinn−1x+n−1m+n∫cosmsinn−2xdx7.\int cos^mx\,sin^nx\,dx=\frac{1}{m+n}cos^{m-1}x\,sin^{n+1}x+\frac{m-1}{m+n}\int cos^{m-2}sin^nx\,dx\\\qquad\qquad\qquad\qquad\:\:\,=-\frac{1}{m+n}cos^{m+1}x\,sin^{n-1}x+\frac{n-1}{m+n}\int cos^msin^{n-2}x\,dx7.∫cosmxsinnxdx=m+n1cosm−1xsinn+1x+m+nm−1∫cosm−2sinnxdx=−m+n1cosm+1xsinn−1x+m+nn−1∫cosmsinn−2xdx

8.∫sin⁡axcos⁡bxdx=−12(a+b)cos⁡(a+b)x−12(a−b)cos⁡(a−b)x+C8.\int \sin ax\cos bx\,dx=-\frac{1}{2(a+b)}\cos\,(a+b)x-\frac{1}{2(a-b)}\cos\,(a-b)x+C8.∫sinaxcosbxdx=−2(a+b)1cos(a+b)x−2(a−b)1cos(a−b)x+C

9.∫sin⁡axsin⁡bxdx=−12(a+b)sin⁡(a+b)x+12(a−b)sin⁡(a−b)x+C9.\int\sin ax\sin bx\,dx=-\frac{1}{2(a+b)}\sin\,(a+b)x+\frac{1}{2(a-b)}\sin\,(a-b)x+C9.∫sinaxsinbxdx=−2(a+b)1sin(a+b)x+2(a−b)1sin(a−b)x+C

10.∫cos⁡axcos⁡bxdx=12(a+b)sin⁡(a+b)x+12(a−b)sin⁡(a−b)x+C10.\int\cos ax\cos bx\,dx=\frac{1}{2(a+b)}\sin\,(a+b)x+\frac{1}{2(a-b)}\sin\,(a-b)x+C10.∫cosaxcosbxdx=2(a+b)1sin(a+b)x+2(a−b)1sin(a−b)x+C

11.∫xsin⁡xdx=sin⁡x−xcos⁡x+C11.\int x\sin{x}dx=\sin{x}-x\cos{x}+C11.∫xsinxdx=sinx−xcosx+C

12.∫xcos⁡xdx=cos⁡x+xsin⁡x+C12.\int x\cos{x}dx=\cos{x}+x\sin{x}+C12.∫xcosxdx=cosx+xsinx+C

13.∫xnsin⁡xdx=−xncos⁡x+n∫xn−1cos⁡xdx13.\int x^n\sin{x}dx=-x^n\cos{x}+n\int x^{n-1}\cos{x}dx13.∫xnsinxdx=−xncosx+n∫xn−1cosxdx

14.∫xncos⁡xdx=xnsin⁡x−n∫xn−1sin⁡xdx14.\int x^n\cos{x}dx=x^n\sin{x}-n\int x^{n-1}\sin{x}dx14.∫xncosxdx=xnsinx−n∫xn−1sinxdx

15.∫dxsinnx=−1n−1⋅cosxsinn−1x+n−2n−1∫dxsinn−2x15.\int\frac{dx}{sin^nx}=-\frac{1}{n-1}·\frac{cos\,x}{sin^{n-1}x}+\frac{n-2}{n-1}\int\frac{dx}{sin^{n-2}x}15.∫sinnxdx=−n−11⋅sinn−1xcosx+n−1n−2∫sinn−2xdx

16.∫dxcosnx=1n−1⋅sinxcosn−1x+n−2n−1∫dxcosn−2x16.\int\frac{dx}{cos^nx}=\frac{1}{n-1}·\frac{sin\,x}{cos^{n-1}x}+\frac{n-2}{n-1}\int\frac{dx}{cos^{n-2}x}16.∫cosnxdx=n−11⋅cosn−1xsinx+n−1n−2∫cosn−2xdx

17.∫dx1±sin⁡x=tan⁡x∓sec⁡x+C17.\int\frac{dx}{1±\sin{x}}=\tan{x}∓\sec{x}+C17.∫1±sinxdx=tanx∓secx+C

18.∫dx1±cos⁡x=−cot⁡x±csc⁡x+C18.\int\frac{dx}{1±\cos{x}}=-\cot{x}±\csc{x}+C18.∫1±cosxdx=−cotx±cscx+C

19.∫dxsin⁡xcos⁡x=ln⁡∣tan⁡x∣+C19.\int\frac{dx}{\sin{x}\cos{x}}=\ln|\tan{x}|+C19.∫sinxcosxdx=ln∣tanx∣+C

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十二.含有tanx,cotx,secx,cscxtan\,x,cot\,x,sec\,x,csc\,xtanx,cotx,secx,cscx的形式

1.∫tan⁡xdx=−ln⁡∣cos⁡x∣+C1.\int\tan{x}dx=-\ln|\cos x|+C1.∫tanxdx=−ln∣cosx∣+C

2.∫tan⁡2xdx=−x+tanx+C2.\int \tan^2{x}dx=-x+tan\,x+C2.∫tan2xdx=−x+tanx+C

3.∫tan⁡nxdx=tan⁡n−1xn−1−∫tan⁡n−2xdx(n≠1)3.\int \tan^n{x}dx=\frac{\tan^{n-1}{x}}{n-1}-\int\tan^{n-2}{x}dx\,(n≠1)3.∫tannxdx=n−1tann−1x−∫tann−2xdx(n=1)

4.∫cotxdx=ln⁡∣sinx∣+C4.\int cot\,x\,dx=\ln|sin\,x|+C4.∫cotxdx=ln∣sinx∣+C

5.∫cot⁡2xdx=−x−cotx+C5.\int\cot^2{x}dx=-x-cot\,x+C5.∫cot2xdx=−x−cotx+C

6.∫cot⁡nxdx=−cot⁡n−1xn−1−∫cot⁡n−2xdx(n≠1)6.\int\cot^n{x}dx=-\frac{\cot^{n-1}{x}}{n-1}-\int\cot^{n-2}{x}dx\,(n≠1)6.∫cotnxdx=−n−1cotn−1x−∫cotn−2xdx(n=1)

7.∫secxdx=ln⁡∣tan(π4+x2)∣+C=ln⁡∣secx+tanx∣+C7.\int sec\,x\,dx=\ln|tan(\frac{\pi}{4}+\frac{x}{2})|+C=\ln|sec\,x+tan\,x|+C7.∫secxdx=ln∣tan(4π+2x)∣+C=ln∣secx+tanx∣+C

8.∫sec2xdx=tanx+C8.\int sec^2x\,dx=tan\,x+C8.∫sec2xdx=tanx+C

9.∫sec⁡nxdx=sec⁡n−2xtan⁡xn−1+n−2n−1∫sec⁡n−2xdx(n≠1)9.\int\sec^n{x}\,dx=\frac{\sec^{n-2}{x}\tan{x}}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}{x}dx\,(n≠1)9.∫secnxdx=n−1secn−2xtanx+n−1n−2∫secn−2xdx(n=1)

10.∫cscxdx=ln⁡∣tanx2∣+C=ln⁡∣cscx−cotx∣+C10.\int csc\,x\,dx=\ln|tan\frac{x}{2}|+C=\ln|csc\,x-cot\,x|+C10.∫cscxdx=ln∣tan2x∣+C=ln∣cscx−cotx∣+C

11.∫csc2xdx=−cotx+C11.\int csc^2x\,dx=-cot\,x+C11.∫csc2xdx=−cotx+C

12.∫csc⁡nxdx=−csc⁡n−2xcot⁡xn−1+n−2n−1∫csc⁡n−2xdx(n≠1)12.\int\csc^n{x}\,dx=-\frac{\csc^{n-2}{x}\cot{x}}{n-1}+\frac{n-2}{n-1}\int\csc^{n-2}{x}dx\,(n≠1)12.∫cscnxdx=−n−1cscn−2xcotx+n−1n−2∫cscn−2xdx(n=1)

13.∫sec⁡xtan⁡xdx=sec⁡x+C13.\int\sec{x}\tan{x}dx=\sec{x}+C13.∫secxtanxdx=secx+C

14.∫csc⁡xcot⁡xdx=−csc⁡x+C14.\int\csc{x}\cot{x}dx=-\csc{x}+C14.∫cscxcotxdx=−cscx+C

15.∫dx1±tan⁡x=12(x±ln⁡∣cos⁡x±sin⁡x∣)+C15.\int\frac{dx}{1±\tan{x}}=\frac{1}{2}(x±\ln|\cos{x}±\sin{x}|)+C15.∫1±tanxdx=21(x±ln∣cosx±sinx∣)+C

16.∫dx1±cot⁡x=12(x∓ln⁡∣sin⁡x±cos⁡x∣)+C16.\int\frac{dx}{1±\cot{x}}=\frac{1}{2}(x∓\ln|\sin{x}±\cos{x}|)+C16.∫1±cotxdx=21(x∓ln∣sinx±cosx∣)+C

17.∫dx1±sec⁡x=x+cot⁡x∓csc⁡x+C17.\int\frac{dx}{1±\sec{x}}=x+\cot{x}∓\csc{x}+C17.∫1±secxdx=x+cotx∓cscx+C

18.∫dx1±cot⁡x=x−tan⁡x±sec⁡x+C18.\int\frac{dx}{1±\cot{x}}=x-\tan{x}±\sec{x}+C18.∫1±cotxdx=x−tanx±secx+C

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十三.含有反三角函数的形式

1.∫arcsin⁡xdx=xarcsin⁡x+1−x2+C1.\int\arcsin{x}dx=x\arcsin{x}+\sqrt{1-x^2}+C1.∫arcsinxdx=xarcsinx+1−x2+C

2.∫arccos⁡xdx=xarccos⁡x−1−x2+C2.\int\arccos{x}dx=x\arccos{x}-\sqrt{1-x^2}+C2.∫arccosxdx=xarccosx−1−x2+C

3.∫arctan⁡xdx=xarctan⁡x−12ln⁡(1+x2)+C3.\int\arctan{x}dx=x\arctan{x}-\frac{1}{2}\ln(1+x^2)+C3.∫arctanxdx=xarctanx−21ln(1+x2)+C

4.∫arccotxdx=xarccotx+12ln⁡(1+x2)+C4.\int\text{arccot}\,xdx=x\text{arccot}\,x+\frac{1}{2}\ln(1+x^2)+C4.∫arccotxdx=xarccotx+21ln(1+x2)+C

5.∫arcsecxdx=xarcsecx−ln⁡∣x+x2−1∣+C5.\int\text{arcsec}\,xdx=x\text{arcsec}\,x-\ln|x+\sqrt{x^2-1}|+C5.∫arcsecxdx=xarcsecx−ln∣x+x2−1∣+C

6.∫arccscxdx=xarccscx+ln⁡∣x+x2−1∣+C6.\int\text{arccsc}\,xdx=x\text{arccsc}\,x+\ln|x+\sqrt{x^2-1}|+C6.∫arccscxdx=xarccscx+ln∣x+x2−1∣+C

7.∫xarcsin⁡xdx=14[x1−x2+(2x2−1)arcsin⁡x]+C7.\int x\arcsin{x}dx=\frac{1}{4}[x\sqrt{1-x^2}+(2x^2-1)\arcsin{x}]+C7.∫xarcsinxdx=41[x1−x2+(2x2−1)arcsinx]+C

8.∫xarccos⁡xdx=14[−x1−x2+(2x2−1)arccos⁡x]+C8.\int x\arccos{x}dx=\frac{1}{4}[-x\sqrt{1-x^2}+(2x^2-1)\arccos{x}]+C8.∫xarccosxdx=41[−x1−x2+(2x2−1)arccosx]+C

9.∫xarctan⁡xdx=12[(1+x2)arctan⁡x−x]+C9.\int x\arctan{x}dx=\frac{1}{2}[(1+x^2)\arctan{x}-x]+C9.∫xarctanxdx=21[(1+x2)arctanx−x]+C

10.∫xarccotxdx=12[(1+x2)arccotx+x]+C10.\int x\text{arccot}\,xdx=\frac{1}{2}[(1+x^2)\text{arccot}\,x+x]+C10.∫xarccotxdx=21[(1+x2)arccotx+x]+C

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十四.含有exe^xex的形式

1.∫axdx=axln⁡a+C1.\int a^xdx=\frac{a^x}{\ln{a}}+C1.∫axdx=lnaax+C

2.∫exdx=ex+C2.\int e^xdx=e^x+C2.∫exdx=ex+C

3.∫xexdx=(x−1)ex+C3.\int xe^xdx=(x-1)e^x+C3.∫xexdx=(x−1)ex+C

4.∫xnexdx=xnex−n∫xn−1exdx+C4.\int x^ne^xdx=x^ne^x-n\int x^{n-1}e^xdx+C4.∫xnexdx=xnex−n∫xn−1exdx+C

5.∫dx1+ex=x−ln⁡(1+ex)+C5.\int\frac{dx}{1+e^x}=x-\ln{(1+e^x)}+C5.∫1+exdx=x−ln(1+ex)+C

6.∫eaxsin⁡bxdx=eaxa2+b2(asin⁡bx−bcos⁡bx)+C6.\int e^{ax}\sin{bx}dx=\frac{e^{ax}}{a^2+b^2}(a\sin{bx}-b\cos{bx})+C6.∫eaxsinbxdx=a2+b2eax(asinbx−bcosbx)+C

7.∫eaxcos⁡bxdx=eaxa2+b2(acos⁡bx+bsin⁡bx)+C7.\int e^{ax}\cos{bx}dx=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})+C7.∫eaxcosbxdx=a2+b2eax(acosbx+bsinbx)+C

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十五.含有lnxln\,xlnx的形式

1.∫ln⁡xdx=x(ln⁡x−1)+C1.\int\ln{x}dx=x(\ln{x}-1)+C1.∫lnxdx=x(lnx−1)+C

2.∫ln⁡xxdx=4x(ln⁡x−1)+C2.\int\frac{\ln{x}}{\sqrt{x}}dx=4\sqrt{x}(\ln{\sqrt{x}-1})+C2.∫xlnxdx=4x(lnx−1)+C

3.∫xln⁡xdx=x24(2ln⁡x−1)+C3.\int x\ln{x}dx=\frac{x^2}{4}(2\ln{x}-1)+C3.∫xlnxdx=4x2(2lnx−1)+C

4.∫xnln⁡xdx=xn+1(n+1)2[(n+1)ln⁡x−1]+C(n≠−1)4.\int x^n\ln{x}dx=\frac{x^{n+1}}{(n+1)^2}[(n+1)\ln{x}-1]+C\,(n≠-1)4.∫xnlnxdx=(n+1)2xn+1[(n+1)lnx−1]+C(n=−1)

5.∫(ln⁡x)2dx=x[(ln⁡x)2−2ln⁡x+2]+C5.\int(\ln{x})^2dx=x[(\ln{x})^2-2\ln{x}+2]+C5.∫(lnx)2dx=x[(lnx)2−2lnx+2]+C

6.∫(ln⁡x)ndx=x(ln⁡x)n−n∫(ln⁡x)n−1dx6.\int(\ln{x})^ndx=x(\ln{x})^n-n\int(\ln{x})^{n-1}dx6.∫(lnx)ndx=x(lnx)n−n∫(lnx)n−1dx

7.∫sin⁡(ln⁡x)dx=x2[sin⁡(ln⁡x)−cos⁡(ln⁡x)]+C7.\int\sin{(\ln{x})}dx=\frac{x}{2}[\sin{(\ln{x})}-\cos{(\ln{x})}]+C7.∫sin(lnx)dx=2x[sin(lnx)−cos(lnx)]+C

8.∫cos⁡(ln⁡x)dx=x2[sin⁡(ln⁡x)+cos⁡(ln⁡x)]+C8.\int\cos{(\ln{x})}dx=\frac{x}{2}[\sin{(\ln{x})}+\cos{(\ln{x})}]+C8.∫cos(lnx)dx=2x[sin(lnx)+cos(lnx)]+C

9.∫ln⁡(x+1+x2)dx=xln⁡(x+1+x2)−1+x2+C9.\int\ln{(x+\sqrt{1+x^2})}dx=x\ln{(x+\sqrt{1+x^2})}-\sqrt{1+x^2}+C9.∫ln(x+1+x2)dx=xln(x+1+x2)−1+x2+C

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十六.含有双曲函数的形式

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十七.常用积分公式

1.狄利克雷积分:

∫0+∞sin⁡xxdx=π2\int_0^{+\infty}\frac{\sin{x}}{x}dx=\frac{\pi}{2}∫0+∞xsinxdx=2π

推广:

∫0+∞e−axsin⁡bxxdx=arctan⁡ba\int_0^{+\infty}e^{-ax}\frac{\sin{bx}}{x}dx=\arctan{\frac{b}{a}}∫0+∞e−axxsinbxdx=arctanab

2.拉普拉斯积分:

∫0+∞cos⁡bxa2+x2dx=π2ae−ab(a,b>0)\int_0^{+\infty}\frac{\cos{bx}}{a^2+x^2}dx=\frac{\pi}{2a}e^{-ab}\,(a,b>0)∫0+∞a2+x2cosbxdx=2aπe−ab(a,b>0)

∫0+∞xsin⁡bxa2+x2dx=π2e−ab(a,b>0)\int_0^{+\infty}\frac{x\sin{bx}}{a^2+x^2}dx=\frac{\pi}{2}e^{-ab}\,(a,b>0)∫0+∞a2+x2xsinbxdx=2πe−ab(a,b>0)

类似的积分:

∫−∞+∞cos⁡xx2+px+qdx=2π4q−p2e−4q−p22cos⁡p2(4q>p2)\int_{-\infty}^{+\infty}\frac{\cos{x}}{x^2+px+q}dx=\frac{2\pi}{\sqrt{4q-p^2}}e^{-\frac{\sqrt{4q-p^2}}{2}}\cos{\frac{p}{2}}\,(4q>p^2)∫−∞+∞x2+px+qcosxdx=4q−p22πe−24q−p2cos2p(4q>p2)

∫−∞+∞sin⁡xx2+px+qdx=−2π4q−p2e−4q−p22sin⁡p2(4q>p2)\int_{-\infty}^{+\infty}\frac{\sin{x}}{x^2+px+q}dx=-\frac{2\pi}{\sqrt{4q-p^2}}e^{-\frac{\sqrt{4q-p^2}}{2}}\sin{\frac{p}{2}}\,(4q>p^2)∫−∞+∞x2+px+qsinxdx=−4q−p22πe−24q−p2sin2p(4q>p2)

3.菲涅尔积分:

∫0+∞sin⁡x2dx=∫0+∞cos⁡x2dx=12π2\int_0^{+\infty}\sin{x^2}dx=\int_0^{+\infty}\cos{x^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}∫0+∞sinx2dx=∫0+∞cosx2dx=212π

推广:

∫0+∞sin⁡xkdx=1kΓ(1k)sin⁡π2k\int_0^{+\infty}\sin{x^k}dx=\frac{1}{k}Γ(\frac{1}{k})\sin{\frac{\pi}{2k}}∫0+∞sinxkdx=k1Γ(k1)sin2kπ

∫0+∞cos⁡xkdx=1kΓ(1k)cos⁡π2k\int_0^{+\infty}\cos{x^k}dx=\frac{1}{k}Γ(\frac{1}{k})\cos{\frac{\pi}{2k}}∫0+∞cosxkdx=k1Γ(k1)cos2kπ

4.泊松积分:

∫0+∞e−ax2cos⁡bxdx=12πae−b24a(a>0)\int_0^{+\infty}e^{-ax^2}\cos{bx}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}}\,(a>0)∫0+∞e−ax2cosbxdx=21aπe−4ab2(a>0)

当b=0,a=1b=0,a=1b=0,a=1时便是欧拉-泊松积分:

∫0+∞e−x2dx=π2\int_0^{+\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}∫0+∞e−x2dx=2π

5.欧拉积分:

∫0+∞xa−11+xdx=πsin⁡aπ(0<a<1)\int_0^{+\infty}\frac{x^{a-1}}{1+x}dx=\frac{\pi}{\sin{a\pi}}\,(0<a<1)∫0+∞1+xxa−1dx=sinaππ(0<a<1)

推论:

∫0+∞dxan+xn=aπnansin⁡πn(a>0,n>1)\int_0^{+\infty}\frac{dx}{a^n+x^n}=\frac{a\pi}{na^n\sin{\frac{\pi}{n}}}\,(a>0,n>1)∫0+∞an+xndx=nansinnπaπ(a>0,n>1)

类似的积分:

P.V.∫0+∞xa−11−xdx=πtan⁡aπ(P.V.P.V.\int_0^{+\infty}\frac{x^{a-1}}{1-x}dx=\frac{\pi}{\tan{a\pi}}(P.V.P.V.∫0+∞1−xxa−1dx=tanaππ(P.V.的含义是柯西主值)))

6.艾哈迈德积分:

∫01arctan⁡x2+2(x2+1)x2+2dx=5π296\int_0^1\frac{\arctan{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}}dx=\frac{5\pi^2}{96}∫01(x2+1)x2+2arctanx2+2dx=965π2

7.考克斯特积分:

∫0π2arccos⁡cos⁡x1+2cos⁡xdx=5π224\int_0^{\frac{\pi}{2}}\arccos{\frac{\cos{x}}{1+2\cos{x}}}dx=\frac{5\pi^2}{24}∫02πarccos1+2cosxcosxdx=245π2

8.伏汝兰尼积分:

∫0+∞f(ax)−f(bx)xdx=[f(+∞)−f(0)]ln⁡ab\int_0^{+\infty}\frac{f(ax)-f(bx)}{x}dx=[f(+\infty)-f(0)]\ln{\frac{a}{b}}∫0+∞xf(ax)−f(bx)dx=[f(+∞)−f(0)]lnba

若lim⁡x→+∞f(x)\displaystyle\lim_{x\to+\infty}{f(x)}x→+∞limf(x)不存在,则

∫0+∞f(ax)−f(bx)xdx=−f(0)ln⁡ab\int_0^{+\infty}\frac{f(ax)-f(bx)}{x}dx=-f(0)\ln{\frac{a}{b}}∫0+∞xf(ax)−f(bx)dx=−f(0)lnba

若lim⁡x→0f(x)\displaystyle\lim_{x\to0}{f(x)}x→0limf(x)不存在,则

∫0+∞f(ax)−f(bx)xdx=f(+∞)ln⁡ab\int_0^{+\infty}\frac{f(ax)-f(bx)}{x}dx=f(+\infty)\ln{\frac{a}{b}}∫0+∞xf(ax)−f(bx)dx=f(+∞)lnba

9.Γ(s)Γ(s)Γ(s)与ζ(s),η(s),β(s)ζ(s),η(s),β(s)ζ(s),η(s),β(s)的乘积的积分表达式:

∫0+∞xs−1ex−1dx=Γ(s)ζ(s)\int_0^{+\infty}\frac{x^{s-1}}{e^x-1}dx=Γ(s)ζ(s)∫0+∞ex−1xs−1dx=Γ(s)ζ(s)

∫0+∞xs−1ex+1dx=Γ(s)η(s)\int_0^{+\infty}\frac{x^{s-1}}{e^x+1}dx=Γ(s)η(s)∫0+∞ex+1xs−1dx=Γ(s)η(s)

∫0+∞xs−1ex+e−xdx=Γ(s)β(s)\int_0^{+\infty}\frac{x^{s-1}}{e^x+e^{-x}}dx=Γ(s)β(s)∫0+∞ex+e−xxs−1dx=Γ(s)β(s)

其中:

Γ(s)=∫0+∞ts−1e−tdtΓ(s)=\int_0^{+\infty}t^{s-1}e^{-t}dtΓ(s)=∫0+∞ts−1e−tdt

ζ(s)=∑n=1∞1nsζ(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}ζ(s)=n=1∑∞ns1

η(s)=∑n=1∞(−1)n+1nsη(s)=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^s}η(s)=n=1∑∞ns(−1)n+1

β(s)=∑n=0∞(−1)n(2n+1)sβ(s)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^s}β(s)=n=0∑∞(2n+1)s(−1)n

10.拉阿伯积分公式:

R(a)=∫aa+1ln⁡Γ(x)dx=a(ln⁡a−1)+ln⁡2πR(a)=\int_a^{a+1}\ln{Γ(x)}dx=a(\ln{a}-1)+\ln{\sqrt{2\pi}}R(a)=∫aa+1lnΓ(x)dx=a(lna−1)+ln2π

11.XXXX:

∫−∞+∞f(x−ax)dx=∫−∞+∞f(x)dx(a>0)\int_{-\infty}^{+\infty}f(x-\frac{a}{x})dx=\int_{-\infty}^{+\infty}f(x)dx\,(a>0)∫−∞+∞f(x−xa)dx=∫−∞+∞f(x)dx(a>0)

12.罗巴切夫斯基积分法:

∫0+∞f(x)sin⁡2xxdx=∫0π2f(x)dx(0≤x<+∞,f(π+x)=f(π−x)=f(x))\int_0^{+\infty}f(x)\frac{\sin^2{x}}{x}dx=\int_0^{\frac{\pi}{2}}f(x)dx\,(0≤x<+\infty,f(\pi+x)=f(\pi-x)=f(x))∫0+∞f(x)xsin2xdx=∫02πf(x)dx(0≤x<+∞,f(π+x)=f(π−x)=f(x))

当f(x)=1f(x)=1f(x)=1时便是狄利克雷积分:

∫0+∞sin⁡xxdx=π2\int_0^{+\infty}\frac{\sin{x}}{x}dx=\frac{\pi}{2}∫0+∞xsinxdx=2π

类似的积分:

∫0+∞f(x)sin⁡2xx2dx=∫0π2f(x)dx(0≤x<+∞,f(π+x)=f(π−x)=f(x))\int_0^{+\infty}f(x)\frac{\sin^2{x}}{x^2}dx=\int_0^{\frac{\pi}{2}}f(x)dx\,(0≤x<+\infty,f(\pi+x)=f(\pi-x)=f(x))∫0+∞f(x)x2sin2xdx=∫02πf(x)dx(0≤x<+∞,f(π+x)=f(π−x)=f(x))

13.三角函数递推公式:

①In=∫sin⁡nxdx①I_n=\int\sin^n{x}dx①In=∫sinnxdx
In=−sin⁡n−1xcos⁡xn+n−1nIn−2\quad I_n=-\frac{\sin^{n-1}{x}\cos{x}}{n}+\frac{n-1}{n}I_{n-2}In=−nsinn−1xcosx+nn−1In−2

②In=∫cos⁡nxdx②I_n=\int\cos^n{x}dx②In=∫cosnxdx
In=sin⁡xcos⁡n−1xn+n−1nIn−2\quad I_n=\frac{\sin{x}\cos^{n-1}{x}}{n}+\frac{n-1}{n}I_{n-2}In=nsinxcosn−1x+nn−1In−2

③∫0π2sin⁡nθdθ=∫0π2cos⁡nθdθ=πΓ(n+12)2Γ(n+22)③\int_0^{\frac{\pi}{2}}\sin^n{θ}dθ=\int_0^{\frac{\pi}{2}}\cos^n{θ}dθ=\frac{\sqrt{\pi}Γ(\frac{n+1}{2})}{2Γ(\frac{n+2}{2})}③∫02πsinnθdθ=∫02πcosnθdθ=2Γ(2n+2)πΓ(2n+1)

④In=∫tan⁡nxdx④I_n=\int\tan^n{x}dx④In=∫tannxdx
In=tan⁡n−1xn−1−In−2\quad I_n=\frac{\tan^{n-1}{x}}{n-1}-I_{n-2}In=n−1tann−1x−In−2

⑤In=∫cot⁡nxdx⑤I_n=\int\cot^n{x}dx⑤In=∫cotnxdx
In=−cot⁡n−1xn−1−In−2\quad I_n=-\frac{\cot^{n-1}{x}}{n-1}-I_{n-2}In=−n−1cotn−1x−In−2

⑥In=∫sec⁡nxdx⑥I_n=\int\sec^n{x}dx⑥In=∫secnxdx
In=sec⁡n−2xtan⁡xn−1+n−2n−1In−2\quad I_n=\frac{\sec^{n-2}{x}\tan{x}}{n-1}+\frac{n-2}{n-1}I_{n-2}In=n−1secn−2xtanx+n−1n−2In−2

⑦In=∫csc⁡nxdx⑦I_n=\int\csc^n{x}dx⑦In=∫cscnxdx
In=−csc⁡n−2xcot⁡xn−1+n−2n−1In−2\quad I_n=-\frac{\csc^{n-2}{x}\cot{x}}{n-1}+\frac{n-2}{n-1}I_{n-2}In=−n−1cscn−2xcotx+n−1n−2In−2

14.B(p,q)B(p,q)B(p,q)的积分表达式:

∫0π2sin⁡pxcos⁡qxdx=12B(p+12,q+12)\int_0^{\frac{\pi}{2}}\sin^p{x}\cos^q{x}dx=\frac{1}{2}B(\frac{p+1}{2},\frac{q+1}{2})∫02πsinpxcosqxdx=21B(2p+1,2q+1)

其中B(p,q)=∫01xp−1(1−x)q−1dx(p,q>0)B(p,q)=\int_0^1x^{p-1}(1-x)^{q-1}dx\,(p,q>0)B(p,q)=∫01xp−1(1−x)q−1dx(p,q>0)

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