- 6. I = lim ⁡ x → 0 x sin ⁡ x 2 − 2 ( 1 − cos ⁡ x ) sin ⁡ x x 4 = I=\lim\limits_{x\to0}\cfrac{x\sin x^2-2(1-\cos x)\sin x}{x^4}= I=x→0limx4xsinx2−2(1−cosx)sinx=______.
- 7. I = lim ⁡ x → 0 ( 1 − cos ⁡ x ) ( 1 − cos ⁡ x 3 ) ⋯ ( 1 − cos ⁡ x n ) ( 1 − cos ⁡ x ) n − 1 = I=\lim\limits_{x\to0}\cfrac{(1-\cos x)(1-\sqrt[3]{\cos x})\cdots(1-\sqrt[n]{\cos x})}{(1-\cos x)^{n-1}}= I=x→0lim(1−cosx)n−1(1−cosx)(1−3cosx )⋯(1−ncosx )=_____.
- 9. I = lim ⁡ x → 0 ∫ x 2 x sin ⁡ ( x t ) t d t x 2 I=\lim\limits_{x\to0}\cfrac{\displaystyle\int^x_{x^2}\cfrac{\sin(xt)}{t}\mathrm{d}t}{x^2} I=x→0limx2∫x2xtsin(xt)dt=_____.
- 14.设 lim ⁡ x → 0 ln ⁡ ( 1 + x + f ( x ) x ) x = 3 \lim\limits_{x\to0}\cfrac{\ln\left(1+x+\cfrac{f(x)}{x}\right)}{x}=3 x→0limxln(1+x+xf(x))=3，则 lim ⁡ x → 0 f ( x ) x 2 \lim\limits_{x\to0}\cfrac{f(x)}{x^2} x→0limx2f(x)=_____.
- 17.设 a , b a,b a,b为常数，且 lim ⁡ x → ∞ ( 1 − x 6 3 − a x 2 − b ) = 0 \lim\limits_{x\to\infty}(\sqrt[3]{1-x^6}-ax^2-b)=0 x→∞lim(31−x6 −ax2−b)=0，则 a = a= a=_____， b = b= b=_____.
- 19.设 x 0 = 0 , x n = 1 + 2 x n − 1 1 + x n − 1 x_0=0,x_n=\cfrac{1+2x_{n-1}}{1+x_{n-1}} x0=0,xn=1+xn−11+2xn−1，则 lim ⁡ n → ∞ x n = \lim\limits_{n\to\infty}x_n= n→∞limxn=_____.
- 30.设 f ( x ) f(x) f(x)在 x = 0 x=0 x=0可导且 f ( 0 ) = 1 , f ′ ( 0 ) = 3 f(0)=1,f'(0)=3 f(0)=1,f′(0)=3，则数列极限 I = lim ⁡ n → ∞ ( f ( 1 n ) ) 1 n 1 − cos ⁡ 1 n = I=\lim\limits_{n\to\infty}\left(f\left(\cfrac{1}{n}\right)\right)^{\frac{\frac{1}{n}}{1-\cos\frac{1}{n}}}= I=n→∞lim(f(n1))1−cosn1n1=_____.
- 31.设 f ( x ) f(x) f(x)在 x = a x=a x=a处二阶导数存在，则 I = lim ⁡ h → 0 f ( a + h ) − f ( a ) h − f ′ ( a ) h = I=\lim\limits_{h\to0}\cfrac{\cfrac{f(a+h)-f(a)}{h}-f'(a)}{h}= I=h→0limhhf(a+h)−f(a)−f′(a)=_____.
- 51.设 ∫ x 2 f ′ ( x ) d x = arcsin ⁡ x + C \displaystyle\int x^2f'(x)\mathrm{d}x=\arcsin x+C ∫x2f′(x)dx=arcsinx+C，则 f ( x ) = f(x)= f(x)=_____.
- 53. I = ∫ x + 1 + 2 ( x + 1 ) 2 − x + 1 d x = I=\displaystyle\int\cfrac{\sqrt{x+1}+2}{(x+1)^2-\sqrt{x+1}}\mathrm{d}x= I=∫(x+1)2−x+1 x+1 +2dx=_____.
- 57. I = ∫ d x x x 4 + 1 d x = I=\displaystyle\int\cfrac{\mathrm{d}x}{x\sqrt{x^4+1}}\mathrm{d}x= I=∫xx4+1 dxdx=_____.
- 60. I = ∫ 0 2 ( x 2 x − x 2 − ( 1 − 1 4 x 2 ) 3 ) d x = I=\displaystyle\int^2_0\left(x\sqrt{2x-x^2}-\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\right)\mathrm{d}x= I=∫02⎝⎛x2x−x2 −(1−41x2)3 ⎠⎞dx=_____.
- 79.当 y > 0 y>0 y>0时，微分方程 ( x − 2 x y − y 2 ) d y + y 2 d x = 0 (x-2xy-y^2)\mathrm{d}y+y^2\mathrm{d}x=0 (x−2xy−y2)dy+y2dx=0的通解为_____。
- 94. 设 z = ∫ 0 1 ∣ x y − t ∣ f ( t ) d t , 0 ⩽ x ⩽ 1 , 0 ⩽ y ⩽ 1 z=\displaystyle\int^1_0|xy-t|f(t)\mathrm{d}t,0\leqslant x\leqslant1,0\leqslant y\leqslant1 z=∫01∣xy−t∣f(t)dt,0⩽x⩽1,0⩽y⩽1，其中 f ( x ) f(x) f(x)为连续函数，则 z x x ′ ′ + z y y ′ ′ = z''_{xx}+z''_{yy}= zxx′′+zyy′′=_____.
写在最后

本节包括了一到一百题的易错题和难题。

6. I = lim ⁡ x → 0 x sin ⁡ x 2 − 2 ( 1 − cos ⁡ x ) sin ⁡ x x 4 = I=\lim\limits_{x\to0}\cfrac{x\sin x^2-2(1-\cos x)\sin x}{x^4}= I=x→0limx4xsinx2−2(1−cosx)sinx=______.

解先做如下变形： I = lim ⁡ x → 0 x sin ⁡ x 2 − 2 sin ⁡ x + 2 sin ⁡ 2 x x 4 . I=\lim\limits_{x\to0}\cfrac{x\sin x^2-2\sin x+2\sin2x}{x^4}. I=x→0limx4xsinx2−2sinx+2sin2x.
可用泰勒展开式
sin ⁡ x = x − 1 6 x 3 + ο ( x 4 ) ⇒ x sin ⁡ x 2 = x ( x 2 + ο ( x 4 ) ) = x 3 + ο ( x 4 ) − 2 sin ⁡ x = − 2 ( x − 1 6 x 3 + ο ( x 4 ) ) = − 2 x + 1 3 x 3 + ο ( x 4 ) sin ⁡ 2 x = 2 x − 1 6 ( 2 x ) 3 + ο ( x 4 ) = 2 x − 4 3 x 3 + ο ( x 4 ) \sin x=x-\cfrac{1}{6}x^3+\omicron(x^4)\\ \Rightarrow x\sin x^2=x(x^2+\omicron(x^4))=x^3+\omicron(x^4)\\ -2\sin x=-2\left(x-\cfrac{1}{6}x^3+\omicron(x^4)\right)=-2x+\cfrac{1}{3}x^3+\omicron(x^4)\\ \sin2x=2x-\cfrac{1}{6}(2x)^3+\omicron(x^4)=2x-\cfrac{4}{3}x^3+\omicron(x^4)\\ sinx=x−61x3+ο(x4)⇒xsinx2=x(x2+ο(x4))=x3+ο(x4)−2sinx=−2(x−61x3+ο(x4))=−2x+31x3+ο(x4)sin2x=2x−61(2x)3+ο(x4)=2x−34x3+ο(x4)
相加得
x sin ⁡ x 2 − 2 sin ⁡ x + 2 sin ⁡ 2 x = 0 + ο ( x 4 ) ( x → 0 ) x\sin x^2-2\sin x+2\sin2x=0+\omicron(x^4)(x\to0) xsinx2−2sinx+2sin2x=0+ο(x4)(x→0)
因此
I = lim ⁡ x → 0 ο ( x 4 ) x 4 = 0. I=\lim\limits_{x\to0}\cfrac{\omicron(x^4)}{x^4}=0. I=x→0limx4ο(x4)=0.
（这道题主要利用了泰勒展开式求解）

7. I = lim ⁡ x → 0 ( 1 − cos ⁡ x ) ( 1 − cos ⁡ x 3 ) ⋯ ( 1 − cos ⁡ x n ) ( 1 − cos ⁡ x ) n − 1 = I=\lim\limits_{x\to0}\cfrac{(1-\cos x)(1-\sqrt[3]{\cos x})\cdots(1-\sqrt[n]{\cos x})}{(1-\cos x)^{n-1}}= I=x→0lim(1−cosx)n−1(1−cosx)(1−3cosx )⋯(1−ncosx )=_____.

解用等价无穷小因子替换：
t ∼ ln ⁡ ( 1 + t ) ( cos ⁡ x ) 1 m − 1 ∼ ln ⁡ [ ( cos ⁡ x ) 1 m − 1 + 1 ] = 1 m ln ⁡ cos ⁡ x . t\sim\ln(1+t)\\ (\cos x)^{\frac{1}{m}}-1\sim\ln[(\cos x)^{\frac{1}{m}}-1+1]=\cfrac{1}{m}\ln\cos x. t∼ln(1+t)(cosx)m1−1∼ln[(cosx)m1−1+1]=m1lncosx.
因此
I = lim ⁡ x → 0 ( cos ⁡ x − 1 ) ( cos ⁡ x 3 − 1 ) ⋯ ( cos ⁡ x n − 1 ) ( cos ⁡ x − 1 ) n − 1 = lim ⁡ x → 0 ln ⁡ cos ⁡ x ln ⁡ cos ⁡ x 3 ⋯ ln ⁡ cos ⁡ x n ( cos ⁡ x − 1 ) n − 1 = 1 n ! lim ⁡ x → 0 ln ⁡ n − 1 cos ⁡ x ( cos ⁡ x − 1 ) n − 1 = 1 n ! . \begin{aligned} I&=\lim\limits_{x\to0}\cfrac{(\cos x-1)(\sqrt[3]{\cos x}-1)\cdots(\sqrt[n]{\cos x}-1)}{(\cos x-1)^{n-1}}\\ &=\lim\limits_{x\to0}\cfrac{\ln\cos x\ln\sqrt[3]{\cos x}\cdots\ln\sqrt[n]{\cos x}}{(\cos x-1)^{n-1}}\\ &=\cfrac{1}{n!}\lim\limits_{x\to0}\cfrac{\ln^{n-1}\cos x}{(\cos x-1)^{n-1}}=\cfrac{1}{n!}. \end{aligned} I=x→0lim(cosx−1)n−1(cosx−1)(3cosx −1)⋯(ncosx −1)=x→0lim(cosx−1)n−1lncosxln3cosx ⋯lnncosx =n!1x→0lim(cosx−1)n−1lnn−1cosx=n!1.
（这道题主要利用了等价无穷小代换求解）

9. I = lim ⁡ x → 0 ∫ x 2 x sin ⁡ ( x t ) t d t x 2 I=\lim\limits_{x\to0}\cfrac{\displaystyle\int^x_{x^2}\cfrac{\sin(xt)}{t}\mathrm{d}t}{x^2} I=x→0limx2∫x2xtsin(xt)dt=_____.

解
∫ x 2 x sin ⁡ ( x t ) t d t = x t = s ∫ x 3 x 2 sin ⁡ s s d s ⇒ I = lim ⁡ x → 0 ∫ x 3 x 2 sin ⁡ s s d s x 2 = lim ⁡ x → 0 sin ⁡ x 2 x 2 − lim ⁡ x → 0 3 sin ⁡ x 3 2 x 2 = lim ⁡ x → 0 sin ⁡ x 2 x 2 − lim ⁡ x → 0 3 2 sin ⁡ x 3 x 2 = 1. \displaystyle\int^x_{x^2}\cfrac{\sin(xt)}{t}\mathrm{d}t\xlongequal{xt=s}\displaystyle\int^{x^2}_{x^3}\cfrac{\sin s}{s}\mathrm{d}s\\ \Rightarrow\begin{aligned} I&=\lim\limits_{x\to0}\cfrac{\displaystyle\int^{x^2}_{x^3}\cfrac{\sin s}{s}\mathrm{d}s}{x^2}=\lim\limits_{x\to0}\cfrac{\sin x^2}{x^2}-\lim\limits_{x\to0}\cfrac{3\sin x^3}{2x^2}\\ &=\lim\limits_{x\to0}\cfrac{\sin x^2}{x^2}-\lim\limits_{x\to0}\cfrac{3}{2}\cfrac{\sin x^3}{x^2}=1. \end{aligned} ∫x2xtsin(xt)dtxt=s ∫x3x2ssinsds⇒I=x→0limx2∫x3x2ssinsds=x→0limx2sinx2−x→0lim2x23sinx3=x→0limx2sinx2−x→0lim23x2sinx3=1.
（这道题主要利用了换元法求解）

14.设 lim ⁡ x → 0 ln ⁡ ( 1 + x + f ( x ) x ) x = 3 \lim\limits_{x\to0}\cfrac{\ln\left(1+x+\cfrac{f(x)}{x}\right)}{x}=3 x→0limxln(1+x+xf(x))=3，则 lim ⁡ x → 0 f ( x ) x 2 \lim\limits_{x\to0}\cfrac{f(x)}{x^2} x→0limx2f(x)=_____.

解由 lim ⁡ x → 0 ln ⁡ ( 1 + x + f ( x ) x ) x = 3 \lim\limits_{x\to0}\cfrac{\ln\left(1+x+\cfrac{f(x)}{x}\right)}{x}=3 x→0limxln(1+x+xf(x))=3，当 x → 0 x\to0 x→0时，分母为无穷小，所以分子也为无穷小，进一步有 lim ⁡ x → 0 ( x + f ( x ) x ) = 0. \lim\limits_{x\to0}\left(x+\cfrac{f(x)}{x}\right)=0. x→0lim(x+xf(x))=0.
因此，当 x → 0 x\to0 x→0时， ln ⁡ ( 1 + x + f ( x ) x ) ∼ x + f ( x ) x \ln\left(1+x+\cfrac{f(x)}{x}\right)\sim x+\cfrac{f(x)}{x} ln(1+x+xf(x))∼x+xf(x)。因此 lim ⁡ x → 0 f ( x ) x 2 = 2. \lim\limits_{x\to0}\cfrac{f(x)}{x^2}=2. x→0limx2f(x)=2.（这道题主要利用了等价无穷小代换求解）

17.设 a , b a,b a,b为常数，且 lim ⁡ x → ∞ ( 1 − x 6 3 − a x 2 − b ) = 0 \lim\limits_{x\to\infty}(\sqrt[3]{1-x^6}-ax^2-b)=0 x→∞lim(31−x6 −ax2−b)=0，则 a = a= a=， b = b= b=.

解由泰勒展开，当 x → ∞ x\to\infty x→∞时，
1 − x 6 3 = − x 2 ( 1 − 1 x 6 ) 1 3 = − x 2 ( 1 − 1 3 x 6 + ο ( x − 6 ) ) . \sqrt[3]{1-x^6}=-x^2\left(1-\cfrac{1}{x^6}\right)^{\frac{1}{3}}=-x^2\left(1-\cfrac{1}{3x^6}+\omicron(x^{-6})\right). 31−x6 =−x2(1−x61)31=−x2(1−3x61+ο(x−6)).
所以 a = − 1 , b = 0. a=-1,b=0. a=−1,b=0.（这道题主要利用了泰勒展开式求解）

19.设 x 0 = 0 , x n = 1 + 2 x n − 1 1 + x n − 1 x_0=0,x_n=\cfrac{1+2x_{n-1}}{1+x_{n-1}} x0=0,xn=1+xn−11+2xn−1，则 lim ⁡ n → ∞ x n = \lim\limits_{n\to\infty}x_n= n→∞limxn=_____.

解显然，
0 < x n = 2 ( 1 + x n − 1 ) − 1 1 + x n − 1 = 2 − 1 1 + x n − 1 < 2 ( n = 1 , 2 , 3 , ⋯ ) 0<x_n=\cfrac{2(1+x_{n-1})-1}{1+x_{n-1}}=2-\cfrac{1}{1+x_{n-1}}<2\qquad(n=1,2,3,\cdots) 0<xn=1+xn−12(1+xn−1)−1=2−1+xn−11<2(n=1,2,3,⋯)
即 x n x_n xn有界。
令 f ( x ) = 2 − 1 1 + x ⇒ x n + 1 = f ( x n ) ( n = 1 , 2 , 3 , ⋯ ) f(x)=2-\cfrac{1}{1+x}\Rightarrow x_{n+1}=f(x_n)(n=1,2,3,\cdots) f(x)=2−1+x1⇒xn+1=f(xn)(n=1,2,3,⋯)单调。
因此 x n x_n xn收敛，记 lim ⁡ n → ∞ x n = a \lim\limits_{n\to\infty}x_n=a n→∞limxn=a。
对递归方程 x n = 1 + 2 x n − 1 1 + x n − 1 x_n=\cfrac{1+2x_{n-1}}{1+x_{n-1}} xn=1+xn−11+2xn−1两边取极限得
a = 1 + 2 a 1 + a . a=\cfrac{1+2a}{1+a}. a=1+a1+2a.
解得 a = 1 + 5 2 . a=\cfrac{1+\sqrt{5}}{2}. a=21+5 .（这道题主要利用了数列极限的方法求解）

30.设 f ( x ) f(x) f(x)在 x = 0 x=0 x=0可导且 f ( 0 ) = 1 , f ′ ( 0 ) = 3 f(0)=1,f'(0)=3 f(0)=1,f′(0)=3，则数列极限 I = lim ⁡ n → ∞ ( f ( 1 n ) ) 1 n 1 − cos ⁡ 1 n = I=\lim\limits_{n\to\infty}\left(f\left(\cfrac{1}{n}\right)\right)^{\frac{\frac{1}{n}}{1-\cos\frac{1}{n}}}= I=n→∞lim(f(n1))1−cosn1n1=_____.

解先化为
I = lim ⁡ n → ∞ e 1 n 1 − cos ⁡ 1 n ln ⁡ f ( 1 n ) I=\lim\limits_{n\to\infty}e^{\frac{\frac{1}{n}}{1-\cos\frac{1}{n}}\ln f(\frac{1}{n})} I=n→∞lime1−cosn1n1lnf(n1)
转化为求
lim ⁡ n → ∞ 1 n 1 − cos ⁡ 1 n ln ⁡ f ( 1 n ) = lim ⁡ n → ∞ 1 n 1 2 1 n 2 [ ln ⁡ f ( 1 n ) − ln ⁡ f ( 0 ) ] = 2 lim ⁡ n → ∞ ln ⁡ f ( 1 n ) − ln ⁡ f ( 0 ) 1 n = 2 ⋅ ( ln ⁡ f ( x ) ) ′ ∣ x = 0 = 2 f ′ ( 0 ) f ( 0 ) = 6. ⇒ I = e 6 . \begin{aligned} \lim\limits_{n\to\infty}\cfrac{\cfrac{1}{n}}{1-\cos\cfrac{1}{n}}\ln f\left(\cfrac{1}{n}\right)&=\lim\limits_{n\to\infty}\cfrac{\cfrac{1}{n}}{\cfrac{1}{2}\cfrac{1}{n^2}}\left[\ln f\left(\cfrac{1}{n}\right)-\ln f(0)\right]\\ &=2\lim\limits_{n\to\infty}\cfrac{\ln f\left(\cfrac{1}{n}\right)-\ln f(0)}{\cfrac{1}{n}}=2\cdot(\ln f(x))'\biggm\vert_{x=0}\\ &=2\cfrac{f'(0)}{f(0)}=6. \end{aligned}\\ \Rightarrow I=e^6. n→∞lim1−cosn1n1lnf(n1)=n→∞lim21n21n1[lnf(n1)−lnf(0)]=2n→∞limn1lnf(n1)−lnf(0)=2⋅(lnf(x))′∣∣∣∣x=0=2f(0)f′(0)=6.⇒I=e6.
（这道题主要利用了对数求极限的方法求解）

31.设 f ( x ) f(x) f(x)在 x = a x=a x=a处二阶导数存在，则 I = lim ⁡ h → 0 f ( a + h ) − f ( a ) h − f ′ ( a ) h = I=\lim\limits_{h\to0}\cfrac{\cfrac{f(a+h)-f(a)}{h}-f'(a)}{h}= I=h→0limhhf(a+h)−f(a)−f′(a)=_____.

解
I = lim ⁡ h → 0 f ( a + h ) − f ( a ) − h f ′ ( a ) h 2 = lim ⁡ h → 0 f ′ ( a + h ) − f ′ ( a ) 2 h = 1 2 f ′ ′ ( a ) . \begin{aligned} I&=\lim\limits_{h\to0}\cfrac{f(a+h)-f(a)-hf'(a)}{h^2}\\ &=\lim\limits_{h\to0}\cfrac{f'(a+h)-f'(a)}{2h}\\ &=\cfrac{1}{2}f''(a). \end{aligned} I=h→0limh2f(a+h)−f(a)−hf′(a)=h→0lim2hf′(a+h)−f′(a)=21f′′(a).
（这道题主要利用了导数的定义求解）

51.设 ∫ x 2 f ′ ( x ) d x = arcsin ⁡ x + C \displaystyle\int x^2f'(x)\mathrm{d}x=\arcsin x+C ∫x2f′(x)dx=arcsinx+C，则 f ( x ) = f(x)= f(x)=_____.

解先求出，有
( ∫ x 2 f ′ ( x ) d x ) ′ = ( arcsin ⁡ x + C ) ′ ⇒ x 2 f ′ ( x ) = 1 1 − x 2 , f ′ ( x ) = 1 x 2 1 − x 2 ⇒ f ( x ) = ∫ 1 x 2 1 − x 2 d x \left(\displaystyle\int x^2f'(x)\mathrm{d}x\right)'=(\arcsin x+C)'\\ \Rightarrow x^2f'(x)=\cfrac{1}{\sqrt{1-x^2}},f'(x)=\cfrac{1}{x^2\sqrt{1-x^2}}\\ \Rightarrow f(x)=\displaystyle\int\cfrac{1}{x^2\sqrt{1-x^2}}\mathrm{d}x (∫x2f′(x)dx)′=(arcsinx+C)′⇒x2f′(x)=1−x2 1,f′(x)=x21−x2 1⇒f(x)=∫x21−x2 1dx
作三角代换 x = sin ⁡ t ( − π 2 < t < t 2 ) x=\sin t\left(-\cfrac{\pi}{2}<t<\cfrac{t}{2}\right) x=sint(−2π<t<2t)得
f ( x ) = ∫ cos ⁡ t sin ⁡ 2 t cos ⁡ t d t = − cot ⁡ x + C = − 1 − x 2 x + C . \begin{aligned} f(x)&=\displaystyle\int\cfrac{\cos t}{\sin^2t\cos t}\mathrm{d}t=-\cot x+C\\ &=-\cfrac{\sqrt{1-x^2}}{x}+C. \end{aligned} f(x)=∫sin2tcostcostdt=−cotx+C=−x1−x2 +C.
（这道题主要利用了三角代换求解）

53. I = ∫ x + 1 + 2 ( x + 1 ) 2 − x + 1 d x = I=\displaystyle\int\cfrac{\sqrt{x+1}+2}{(x+1)^2-\sqrt{x+1}}\mathrm{d}x= I=∫(x+1)2−x+1 x+1 +2dx=_____.

解令 x + 1 = t \sqrt{x+1}=t x+1 =t，则 x = t 2 − 1 , d x = 2 t d t x=t^2-1,\mathrm{d}x=2t\mathrm{d}t x=t2−1,dx=2tdt，于是
I = ∫ t + 2 t 4 − t 2 t d t = 2 ∫ t + 2 t 3 − 1 d t = 2 ∫ ( t 2 + t + 1 ) − ( t 2 − 1 ) ( t − 1 ) ( t 2 + t + 1 ) d t = 2 ∫ ( 1 t − 1 − t + 1 t 2 + t + 1 ) d t = 2 ∫ ( 1 t − 1 − 1 2 2 t + 1 + 1 t 2 + t + 1 ) d t = 2 ∫ 1 t − 1 d t − ∫ d ( t 2 + t + 1 ) t 2 + t + 1 − ∫ d ( t + 1 2 ) ( t + 1 2 ) 2 + ( 3 2 ) 2 = 2 ln ⁡ ∣ t − 1 ∣ − ln ⁡ ∣ t 2 + t + 1 ∣ − 2 3 arctan ⁡ 2 x + 1 + 1 3 + C . \begin{aligned} I&=\displaystyle\int\cfrac{t+2}{t^4-t}2t\mathrm{d}t=2\displaystyle\int\cfrac{t+2}{t^3-1}\mathrm{d}t=2\displaystyle\int\cfrac{(t^2+t+1)-(t^2-1)}{(t-1)(t^2+t+1)}\mathrm{d}t\\ &=2\displaystyle\int\left(\cfrac{1}{t-1}-\cfrac{t+1}{t^2+t+1}\right)\mathrm{d}t=2\displaystyle\int\left(\cfrac{1}{t-1}-\cfrac{1}{2}\cfrac{2t+1+1}{t^2+t+1}\right)\mathrm{d}t\\ &=2\displaystyle\int\cfrac{1}{t-1}\mathrm{d}t-\displaystyle\int\cfrac{\mathrm{d}(t^2+t+1)}{t^2+t+1}-\displaystyle\int\cfrac{\mathrm{d}\left(t+\cfrac{1}{2}\right)}{\left(t+\cfrac{1}{2}\right)^2+\left(\cfrac{\sqrt{3}}{2}\right)^2}\\ &=2\ln|t-1|-\ln|t^2+t+1|-\cfrac{2}{\sqrt{3}}\arctan\cfrac{2\sqrt{x+1}+1}{\sqrt{3}}+C. \end{aligned} I=∫t4−tt+22tdt=2∫t3−1t+2dt=2∫(t−1)(t2+t+1)(t2+t+1)−(t2−1)dt=2∫(t−11−t2+t+1t+1)dt=2∫(t−11−21t2+t+12t+1+1)dt=2∫t−11dt−∫t2+t+1d(t2+t+1)−∫(t+21)2+(23 )2d(t+21)=2ln∣t−1∣−ln∣t2+t+1∣−3 2arctan3 2x+1 +1+C.
（这道题主要利用了换元法求解）

57. I = ∫ d x x x 4 + 1 d x = I=\displaystyle\int\cfrac{\mathrm{d}x}{x\sqrt{x^4+1}}\mathrm{d}x= I=∫xx4+1 dxdx=_____.

解
I = ∫ d x x 3 1 x 4 + 1 d x = − 1 2 ∫ [ 1 + ( 1 x 2 ) 2 ] − 1 2 d ( 1 x 2 ) = − 1 2 ln ⁡ ∣ 1 x 2 + 1 + 1 x 4 ∣ + C . \begin{aligned} I&=\displaystyle\int\cfrac{\mathrm{d}x}{x^3\sqrt{\cfrac{1}{x^4}+1}}\mathrm{d}x=\cfrac{-1}{2}\displaystyle\int\left[1+\left(\cfrac{1}{x^2}\right)^2\right]^{-\frac{1}{2}}\mathrm{d}\left(\cfrac{1}{x^2}\right)\\ &=-\cfrac{1}{2}\ln\left|\cfrac{1}{x^2}+\sqrt{1+\cfrac{1}{x^4}}\right|+C. \end{aligned} I=∫x3x41+1 dxdx=2−1∫⎣⎡1+(x21)2⎦⎤−21d(x21)=−21ln∣∣∣∣∣∣x21+1+x41 ∣∣∣∣∣∣+C.
（这道题主要利用了分部积分法求解）

60. I = ∫ 0 2 ( x 2 x − x 2 − ( 1 − 1 4 x 2 ) 3 ) d x = I=\displaystyle\int^2_0\left(x\sqrt{2x-x^2}-\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\right)\mathrm{d}x= I=∫02⎝⎛x2x−x2 −(1−41x2)3 ⎠⎞dx=_____.

解
I = ∫ 0 2 x 2 x − x 2 d x − ∫ 0 2 ( 1 − 1 4 x 2 ) 3 d x . I=\displaystyle\int^2_0x\sqrt{2x-x^2}\mathrm{d}x-\displaystyle\int^2_0\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\mathrm{d}x. I=∫02x2x−x2 dx−∫02(1−41x2)3 dx.
而
∫ 0 2 x 2 x − x 2 d x = ∫ 0 2 x 1 − ( x − 1 ) 2 d x = 令 x − 1 = t ∫ − 1 1 t 1 − t 2 d t + ∫ − 1 1 1 − t 2 d t = 0 + 2 ∫ − 1 0 1 − t 2 d t = 1 2 π . ∫ 0 2 ( 1 − 1 4 x 2 ) 3 d x = 令 x = 2 sin ⁡ t 2 ∫ 0 π 2 cos ⁡ 4 t d t = 3 8 π . I = π 8 . \begin{aligned} \displaystyle\int^2_0x\sqrt{2x-x^2}\mathrm{d}x&=\displaystyle\int^2_0x\sqrt{1-(x-1)^2}\mathrm{d}x\\ &\xlongequal{\text{令}x-1=t}\displaystyle\int^1_{-1}t\sqrt{1-t^2}\mathrm{d}t+\displaystyle\int^1_{-1}\sqrt{1-t^2}\mathrm{d}t\\ &=0+2\displaystyle\int^0_{-1}\sqrt{1-t^2}\mathrm{d}t=\cfrac{1}{2}\pi. \end{aligned}\\ \begin{aligned} \displaystyle\int^2_0\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\mathrm{d}x\xlongequal{\text{令}x=2\sin t}2\displaystyle\int^{\frac{\pi}{2}}_0\cos^4t\mathrm{d}t=\cfrac{3}{8}\pi. \end{aligned}\\ I=\cfrac{\pi}{8}. ∫02x2x−x2 dx=∫02x1−(x−1)2 dx令x−1=t ∫−11t1−t2 dt+∫−111−t2 dt=0+2∫−101−t2 dt=21π.∫02(1−41x2)3 dx令x=2sint 2∫02πcos4tdt=83π.I=8π.
（这道题主要利用了多次换元积分的方法求解）

79.当 y > 0 y>0 y>0时，微分方程 ( x − 2 x y − y 2 ) d y + y 2 d x = 0 (x-2xy-y^2)\mathrm{d}y+y^2\mathrm{d}x=0 (x−2xy−y2)dy+y2dx=0的通解为_____。

解当 y > 0 y>0 y>0时，方程可改写为 d x d y + ( 1 y 2 − 2 y ) x = 1 \cfrac{\mathrm{d}x}{\mathrm{d}y}+\left(\cfrac{1}{y^2}-\cfrac{2}{y}\right)x=1 dydx+(y21−y2)x=1。
两边乘 μ ( y ) = e ∫ ( 1 y 2 − 1 y ) d y = e − 1 y − 2 ln ⁡ y = 1 y 2 e − 1 y \mu(y)=e^{\int(\frac{1}{y^2}-\frac{1}{y})\mathrm{d}y}=e^{-\frac{1}{y}-2\ln y}=\cfrac{1}{y^2}e^{-\frac{1}{y}} μ(y)=e∫(y21−y1)dy=e−y1−2lny=y21e−y1得
d d y ( 1 y 2 e − 1 y x ) = 1 y 2 e − 1 y . \cfrac{\mathrm{d}}{\mathrm{d}y}\left(\cfrac{1}{y^2}e^{-\frac{1}{y}}x\right)=\cfrac{1}{y^2}e^{-\frac{1}{y}}. dyd(y21e−y1x)=y21e−y1.
积分得
1 y 2 e − 1 y x = ∫ 1 y 2 e − 1 y d y + C = e − 1 y + C . \cfrac{1}{y^2}e^{-\frac{1}{y}}x=\displaystyle\int\cfrac{1}{y^2}e^{-\frac{1}{y}}\mathrm{d}y+C=e^{-\frac{1}{y}}+C. y21e−y1x=∫y21e−y1dy+C=e−y1+C.
通解为
x = y 2 e 1 y ( C + e − 1 y ) = y 2 ( C e 1 y + 1 ) . x=y^2e^{\frac{1}{y}}(C+e^{-\frac{1}{y}})=y^2(Ce^{\frac{1}{y}}+1). x=y2ey1(C+e−y1)=y2(Cey1+1).
（这道题主要利用了分离参数的方法求解）

94. 设 z = ∫ 0 1 ∣ x y − t ∣ f ( t ) d t , 0 ⩽ x ⩽ 1 , 0 ⩽ y ⩽ 1 z=\displaystyle\int^1_0|xy-t|f(t)\mathrm{d}t,0\leqslant x\leqslant1,0\leqslant y\leqslant1 z=∫01∣xy−t∣f(t)dt,0⩽x⩽1,0⩽y⩽1，其中 f ( x ) f(x) f(x)为连续函数，则 z x x ′ ′ + z y y ′ ′ = z''_{xx}+z''_{yy}= zxx′′+zyy′′=_____.

解
z = ∫ 0 1 ∣ x y − t ∣ f ( t ) d t = ∫ 0 x y ( x y − t ) f ( t ) d t + ∫ x y 1 ( t − x y ) f ( t ) d t = x y ∫ 0 x y f ( t ) d t − ∫ 0 x y t f ( t ) d t + ∫ x y 1 t f ( t ) d t − x y ∫ x y 1 f ( t ) d t . \begin{aligned} z&=\displaystyle\int^1_0|xy-t|f(t)\mathrm{d}t\\ &=\displaystyle\int^{xy}_0(xy-t)f(t)\mathrm{d}t+\displaystyle\int^1_{xy}(t-xy)f(t)\mathrm{d}t\\ &=xy\displaystyle\int^{xy}_0f(t)\mathrm{d}t-\displaystyle\int^{xy}_0tf(t)\mathrm{d}t+\displaystyle\int^1_{xy}tf(t)\mathrm{d}t-xy\displaystyle\int^1_{xy}f(t)\mathrm{d}t. \end{aligned} z=∫01∣xy−t∣f(t)dt=∫0xy(xy−t)f(t)dt+∫xy1(t−xy)f(t)dt=xy∫0xyf(t)dt−∫0xytf(t)dt+∫xy1tf(t)dt−xy∫xy1f(t)dt.
则
z x ′ = y ∫ 0 x y f ( t ) d t + x y 2 f ( x y ) − x y 2 f ( x y ) − x y 2 f ( x y ) − y ∫ x y 1 f ( t ) d t + x y 2 f ( x y ) = y ∫ 0 x y f ( t ) d t − y ∫ x y 1 f ( t ) d t . z x x ′ ′ = y 2 f ( x y ) + y 2 f ( x y ) = 2 y 2 f ( x y ) . \begin{aligned} z_x'&=y\displaystyle\int^{xy}_0f(t)\mathrm{d}t+xy^2f(xy)-xy^2f(xy)-xy^2f(xy)-y\displaystyle\int^1_{xy}f(t)\mathrm{d}t+xy^2f(xy)\\ &=y\displaystyle\int^{xy}_0f(t)\mathrm{d}t-y\displaystyle\int^1_{xy}f(t)\mathrm{d}t. \end{aligned}\\ z''_{xx}=y^2f(xy)+y^2f(xy)=2y^2f(xy). zx′=y∫0xyf(t)dt+xy2f(xy)−xy2f(xy)−xy2f(xy)−y∫xy1f(t)dt+xy2f(xy)=y∫0xyf(t)dt−y∫xy1f(t)dt.zxx′′=y2f(xy)+y2f(xy)=2y2f(xy).
由变量的对称性可知
z y y ′ ′ = 2 x 2 f ( x y ) . z''_{yy}=2x^2f(xy). zyy′′=2x2f(xy).
则 z x x ′ ′ + z y y ′ ′ = 2 ( x 2 + y 2 ) f ( x y ) z''_{xx}+z''_{yy}=2(x^2+y^2)f(xy) zxx′′+zyy′′=2(x2+y2)f(xy)。（这道题主要利用了多元函数求导的方法求解）

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李永乐数学基础过关660题高等数学填空题

目录

6. I = lim ⁡ x → 0 x sin ⁡ x 2 − 2 ( 1 − cos ⁡ x ) sin ⁡ x x 4 = I=\lim\limits_{x\to0}\cfrac{x\sin x^2-2(1-\cos x)\sin x}{x^4}= I=x→0limx4xsinx2−2(1−cosx)sinx=______.

7. I = lim ⁡ x → 0 ( 1 − cos ⁡ x ) ( 1 − cos ⁡ x 3 ) ⋯ ( 1 − cos ⁡ x n ) ( 1 − cos ⁡ x ) n − 1 = I=\lim\limits_{x\to0}\cfrac{(1-\cos x)(1-\sqrt[3]{\cos x})\cdots(1-\sqrt[n]{\cos x})}{(1-\cos x)^{n-1}}= I=x→0lim(1−cosx)n−1(1−cosx)(1−3cosx )⋯(1−ncosx )=_____.

9. I = lim ⁡ x → 0 ∫ x 2 x sin ⁡ ( x t ) t d t x 2 I=\lim\limits_{x\to0}\cfrac{\displaystyle\int^x_{x^2}\cfrac{\sin(xt)}{t}\mathrm{d}t}{x^2} I=x→0limx2∫x2xtsin(xt)dt=_____.

14.设 lim ⁡ x → 0 ln ⁡ ( 1 + x + f ( x ) x ) x = 3 \lim\limits_{x\to0}\cfrac{\ln\left(1+x+\cfrac{f(x)}{x}\right)}{x}=3 x→0limxln(1+x+xf(x))=3，则 lim ⁡ x → 0 f ( x ) x 2 \lim\limits_{x\to0}\cfrac{f(x)}{x^2} x→0limx2f(x)=_____.

17.设 a , b a,b a,b为常数，且 lim ⁡ x → ∞ ( 1 − x 6 3 − a x 2 − b ) = 0 \lim\limits_{x\to\infty}(\sqrt[3]{1-x^6}-ax^2-b)=0 x→∞lim(31−x6 −ax2−b)=0，则 a = a= a=， b = b= b=.

19.设 x 0 = 0 , x n = 1 + 2 x n − 1 1 + x n − 1 x_0=0,x_n=\cfrac{1+2x_{n-1}}{1+x_{n-1}} x0=0,xn=1+xn−11+2xn−1，则 lim ⁡ n → ∞ x n = \lim\limits_{n\to\infty}x_n= n→∞limxn=_____.

31.设 f ( x ) f(x) f(x)在 x = a x=a x=a处二阶导数存在，则 I = lim ⁡ h → 0 f ( a + h ) − f ( a ) h − f ′ ( a ) h = I=\lim\limits_{h\to0}\cfrac{\cfrac{f(a+h)-f(a)}{h}-f'(a)}{h}= I=h→0limhhf(a+h)−f(a)−f′(a)=_____.

51.设 ∫ x 2 f ′ ( x ) d x = arcsin ⁡ x + C \displaystyle\int x^2f'(x)\mathrm{d}x=\arcsin x+C ∫x2f′(x)dx=arcsinx+C，则 f ( x ) = f(x)= f(x)=_____.

53. I = ∫ x + 1 + 2 ( x + 1 ) 2 − x + 1 d x = I=\displaystyle\int\cfrac{\sqrt{x+1}+2}{(x+1)^2-\sqrt{x+1}}\mathrm{d}x= I=∫(x+1)2−x+1 x+1 +2dx=_____.

57. I = ∫ d x x x 4 + 1 d x = I=\displaystyle\int\cfrac{\mathrm{d}x}{x\sqrt{x^4+1}}\mathrm{d}x= I=∫xx4+1 dxdx=_____.

60. I = ∫ 0 2 ( x 2 x − x 2 − ( 1 − 1 4 x 2 ) 3 ) d x = I=\displaystyle\int^2_0\left(x\sqrt{2x-x^2}-\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\right)\mathrm{d}x= I=∫02⎝⎛x2x−x2 −(1−41x2)3 ⎠⎞dx=_____.

79.当 y > 0 y>0 y>0时，微分方程 ( x − 2 x y − y 2 ) d y + y 2 d x = 0 (x-2xy-y^2)\mathrm{d}y+y^2\mathrm{d}x=0 (x−2xy−y2)dy+y2dx=0的通解为_____。

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李永乐数学基础过关660题高等数学填空题

目录

6. I = lim ⁡ x → 0 x sin ⁡ x 2 − 2 ( 1 − cos ⁡ x ) sin ⁡ x x 4 = I=\lim\limits_{x\to0}\cfrac{x\sin x^2-2(1-\cos x)\sin x}{x^4}= I=x→0lim​x4xsinx2−2(1−cosx)sinx​=______.

9. I = lim ⁡ x → 0 ∫ x 2 x sin ⁡ ( x t ) t d t x 2 I=\lim\limits_{x\to0}\cfrac{\displaystyle\int^x_{x^2}\cfrac{\sin(xt)}{t}\mathrm{d}t}{x^2} I=x→0lim​x2∫x2x​tsin(xt)​dt​=_____.

14.设 lim ⁡ x → 0 ln ⁡ ( 1 + x + f ( x ) x ) x = 3 \lim\limits_{x\to0}\cfrac{\ln\left(1+x+\cfrac{f(x)}{x}\right)}{x}=3 x→0lim​xln(1+x+xf(x)​)​=3，则 lim ⁡ x → 0 f ( x ) x 2 \lim\limits_{x\to0}\cfrac{f(x)}{x^2} x→0lim​x2f(x)​=_____.

17.设 a , b a,b a,b为常数，且 lim ⁡ x → ∞ ( 1 − x 6 3 − a x 2 − b ) = 0 \lim\limits_{x\to\infty}(\sqrt[3]{1-x^6}-ax^2-b)=0 x→∞lim​(31−x6 ​−ax2−b)=0，则 a = a= a=， b = b= b=.

19.设 x 0 = 0 , x n = 1 + 2 x n − 1 1 + x n − 1 x_0=0,x_n=\cfrac{1+2x_{n-1}}{1+x_{n-1}} x0​=0,xn​=1+xn−1​1+2xn−1​​，则 lim ⁡ n → ∞ x n = \lim\limits_{n\to\infty}x_n= n→∞lim​xn​=_____.

31.设 f ( x ) f(x) f(x)在 x = a x=a x=a处二阶导数存在，则 I = lim ⁡ h → 0 f ( a + h ) − f ( a ) h − f ′ ( a ) h = I=\lim\limits_{h\to0}\cfrac{\cfrac{f(a+h)-f(a)}{h}-f'(a)}{h}= I=h→0lim​hhf(a+h)−f(a)​−f′(a)​=_____.

51.设 ∫ x 2 f ′ ( x ) d x = arcsin ⁡ x + C \displaystyle\int x^2f'(x)\mathrm{d}x=\arcsin x+C ∫x2f′(x)dx=arcsinx+C，则 f ( x ) = f(x)= f(x)=_____.

53. I = ∫ x + 1 + 2 ( x + 1 ) 2 − x + 1 d x = I=\displaystyle\int\cfrac{\sqrt{x+1}+2}{(x+1)^2-\sqrt{x+1}}\mathrm{d}x= I=∫(x+1)2−x+1 ​x+1 ​+2​dx=_____.

57. I = ∫ d x x x 4 + 1 d x = I=\displaystyle\int\cfrac{\mathrm{d}x}{x\sqrt{x^4+1}}\mathrm{d}x= I=∫xx4+1 ​dx​dx=_____.

60. I = ∫ 0 2 ( x 2 x − x 2 − ( 1 − 1 4 x 2 ) 3 ) d x = I=\displaystyle\int^2_0\left(x\sqrt{2x-x^2}-\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\right)\mathrm{d}x= I=∫02​⎝⎛​x2x−x2 ​−(1−41​x2)3 ​⎠⎞​dx=_____.

79.当 y > 0 y>0 y>0时，微分方程 ( x − 2 x y − y 2 ) d y + y 2 d x = 0 (x-2xy-y^2)\mathrm{d}y+y^2\mathrm{d}x=0 (x−2xy−y2)dy+y2dx=0的通解为_____。

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6. I = lim ⁡ x → 0 x sin ⁡ x 2 − 2 ( 1 − cos ⁡ x ) sin ⁡ x x 4 = I=\lim\limits_{x\to0}\cfrac{x\sin x^2-2(1-\cos x)\sin x}{x^4}= I=x→0limx4xsinx2−2(1−cosx)sinx=______.

9. I = lim ⁡ x → 0 ∫ x 2 x sin ⁡ ( x t ) t d t x 2 I=\lim\limits_{x\to0}\cfrac{\displaystyle\int^x_{x^2}\cfrac{\sin(xt)}{t}\mathrm{d}t}{x^2} I=x→0limx2∫x2xtsin(xt)dt=_____.

14.设 lim ⁡ x → 0 ln ⁡ ( 1 + x + f ( x ) x ) x = 3 \lim\limits_{x\to0}\cfrac{\ln\left(1+x+\cfrac{f(x)}{x}\right)}{x}=3 x→0limxln(1+x+xf(x))=3，则 lim ⁡ x → 0 f ( x ) x 2 \lim\limits_{x\to0}\cfrac{f(x)}{x^2} x→0limx2f(x)=_____.

17.设 a , b a,b a,b为常数，且 lim ⁡ x → ∞ ( 1 − x 6 3 − a x 2 − b ) = 0 \lim\limits_{x\to\infty}(\sqrt[3]{1-x^6}-ax^2-b)=0 x→∞lim(31−x6 −ax2−b)=0，则 a = a= a=， b = b= b=.

19.设 x 0 = 0 , x n = 1 + 2 x n − 1 1 + x n − 1 x_0=0,x_n=\cfrac{1+2x_{n-1}}{1+x_{n-1}} x0=0,xn=1+xn−11+2xn−1，则 lim ⁡ n → ∞ x n = \lim\limits_{n\to\infty}x_n= n→∞limxn=_____.

31.设 f ( x ) f(x) f(x)在 x = a x=a x=a处二阶导数存在，则 I = lim ⁡ h → 0 f ( a + h ) − f ( a ) h − f ′ ( a ) h = I=\lim\limits_{h\to0}\cfrac{\cfrac{f(a+h)-f(a)}{h}-f'(a)}{h}= I=h→0limhhf(a+h)−f(a)−f′(a)=_____.

53. I = ∫ x + 1 + 2 ( x + 1 ) 2 − x + 1 d x = I=\displaystyle\int\cfrac{\sqrt{x+1}+2}{(x+1)^2-\sqrt{x+1}}\mathrm{d}x= I=∫(x+1)2−x+1 x+1 +2dx=_____.

57. I = ∫ d x x x 4 + 1 d x = I=\displaystyle\int\cfrac{\mathrm{d}x}{x\sqrt{x^4+1}}\mathrm{d}x= I=∫xx4+1 dxdx=_____.

60. I = ∫ 0 2 ( x 2 x − x 2 − ( 1 − 1 4 x 2 ) 3 ) d x = I=\displaystyle\int^2_0\left(x\sqrt{2x-x^2}-\sqrt{\left(1-\cfrac{1}{4}x^2\right)^3}\right)\mathrm{d}x= I=∫02⎝⎛x2x−x2 −(1−41x2)3 ⎠⎞dx=_____.