1. 知识归纳：

1.1 定积分性质

以下结论在 f ( x ) , g ( x ) , ∣ f ( x ) ∣ , f\left( x \right),g\left( x \right),\left| {f\left( x \right)} \right|, f(x),g(x),∣f(x)∣, 在 [ a , b ] [ a , c ] [ c , b ] \left[ {a,b} \right]\left[ {a,c} \right]\left[ {c,b} \right] [a,b][a,c][c,b] 可积，且 a < c < b a<c<b a<c<b

(线性性质) ∫ a b [ α f ( x ) ± β g ( x ) ] d x = α ∫ a b f ( x ) d x ± β ∫ a b g ( x ) d x ( α β ∈ R ) \int_a^b {\left[ {\alpha f\left( x \right) \pm \beta g\left( x \right)} \right]dx} = \alpha \int_a^b {f\left( x \right)} dx \pm \beta \int_a^b {g\left( x \right)} dx\left( {\alpha \beta \in R} \right) ∫ab[αf(x)±βg(x)]dx=α∫abf(x)dx±β∫abg(x)dx(αβ∈R)
(非负性质) f ( x ) f\left( x \right) f(x) 在 [ a , b ] \left[ {a,b} \right] [a,b] 上非负可积，则 ∫ a b f ( x ) d x ≥ 0 \int_a^b {f\left(x \right)} dx \ge {\rm{0}} ∫abf(x)dx≥0
(单调性)若 f ( x ) ≥ g ( x ) ( x ∈ [ a , b ] ) f\left( x \right) \ge g\left( x \right)\left( {x \in \left[ {a,b} \right]} \right) f(x)≥g(x)(x∈[a,b])，则 ∫ a b f ( x ) d x ≥ ∫ a b g ( x ) d x \int_a^b {f\left( x \right)} dx \ge \int_a^b {g\left( x \right)} dx ∫abf(x)dx≥∫abg(x)dx
若 m ≤ f ( x ) ≤ M ( x ∈ [ a , b ] ) , m ( b − a ) ≤ ∫ a b f ( x ) d x ≤ M ( b − a ) m \le f\left( x \right) \le M\left( {x \in \left[ {a,b} \right]} \right),m\left( {b - a} \right) \le \int_a^b {f\left( x \right)} dx \le M\left( {b - a} \right) m≤f(x)≤M(x∈[a,b]),m(b−a)≤∫abf(x)dx≤M(b−a)
∣ ∫ a b f ( x ) d x ∣ ≤ ∫ a b ∣ f ( x ) ∣ d x \left| {\int_a^b {f\left( x \right)} dx} \right| \le \int_a^b {\left| {f\left( x \right)} \right|} dx ∣∣∣∫abf(x)dx∣∣∣≤∫ab∣f(x)∣dx
(区间可加性) ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int_a^b {f\left( x \right)} dx{\rm{ = }}\int_a^c {f\left( x \right)} dx{\rm{ + }}\int_c^b {f\left( x \right)} dx ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx
(积分中值定理) ∫ a b f ( x ) g ( x ) d x = f ( ξ ) ∫ a b g ( x ) d x \int_a^b {f\left( x \right)} g\left( x \right)dx{\rm{ = }}f\left( \xi \right)\int_a^b {g\left( x \right)} dx ∫abf(x)g(x)dx=f(ξ)∫abg(x)dx 推论：令 g ( x ) = 1 ∫ a b f ( x ) d x = f ( ξ ) ( b − a ) g\left( x \right){\rm{ = 1}}\int_a^b {f\left( x \right)} dx{\rm{ = }}f\left( \xi \right)\left( {b - a} \right) g(x)=1∫abf(x)dx=f(ξ)(b−a), ( 积分中值： ∫ a b f ( x ) d x ( b − a ) ) \left(积分中值：{\frac{{\int_a^b {f\left( x \right)} dx}}{{\left( {b - a} \right)}}} \right) (积分中值：(b−a)∫abf(x)dx)
C a u c h y − S c h w a r t z 不等式 : ( ∫ a b f ( x ) g ( x ) d x ) 2 ≤ ∫ a b f 2 ( x ) d x ∫ a b g 2 ( x ) d x ( 竞赛必记 ) M i n k o w w s k i 不等式 : ( ∫ a b [ f ( x ) + g ( x ) ] 2 d x ) 1 2 ≤ ( ∫ a b f 2 ( x ) d x ) 1 2 + ( ∫ a b g 2 ( x ) d x ) 1 2 \begin{array}{l} Cauchy - Schwartz不等式:{\left( {\int_a^b {f\left( x \right)} g\left( x \right)dx} \right)^2} \le \int_a^b {{f^2}\left( x \right)} dx\int_a^b {{g^2}\left( x \right)} dx\left( {竞赛必记} \right)\\ Minkowwski不等式:{\left( {\int_a^b {{{\left[ {f\left( x \right) + g\left( x \right)} \right]}^2}dx} } \right)^{\frac{1}{2}}} \le {\left( {\int_a^b {{f^2}\left( x \right)} dx} \right)^{\frac{1}{2}}} + {\left( {\int_a^b {{g^2}\left( x \right)} dx} \right)^{\frac{1}{2}}} \end{array} Cauchy−Schwartz不等式:(∫abf(x)g(x)dx)2≤∫abf2(x)dx∫abg2(x)dx(竞赛必记)Minkowwski不等式:(∫ab[f(x)+g(x)]2dx)21≤(∫abf2(x)dx)21+(∫abg2(x)dx)21
变上下限积分求导：一般，若 u ( x ) , v ( x ) u\left( x \right),v\left( x \right) u(x),v(x) ，是可微函数， f f f 是连续函数，则有： d d x ( ∫ v ( x ) u ( x ) f ( t ) d t ) = f ( u ( x ) ) u ′ ( x ) − f ( v ( x ) ) v ′ ( x ) \frac{d}{{dx}}\left( {\int_{v\left( x \right)}^{u\left( x \right)} {f\left( t \right)dt} } \right)=f(u(x))u'(x)-f(v(x))v'(x) dxd(∫v(x)u(x)f(t)dt)=f(u(x))u′(x)−f(v(x))v′(x)

1.2 计算技巧公式：

f ( x ) f(x) f(x) 奇函数，则 ∫ − a a f ( x ) d x = 0 \int_{ - a}^a {f\left( x \right)} dx = 0 ∫−aaf(x)dx=0
f ( x ) f(x) f(x) 偶函数，则 ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x \int_{ - a}^a {f\left( x \right)} dx = 2\int_0^a {f\left( x \right)} dx ∫−aaf(x)dx=2∫0af(x)dx
f ( x ) f(x) f(x) 周期函数 l l l 函数，则 ∫ a a + l f ( x ) d x = ∫ 0 l f ( x ) d x = ∫ − l 2 l 2 f ( x ) d x \int_a^{a + l} {f\left( x \right)} dx = \int_0^l {f\left( x \right)} dx = \int_{ - \frac{l}{2}}^{\frac{l}{2}} {f\left( x \right)} dx ∫aa+lf(x)dx=∫0lf(x)dx=∫−2l2lf(x)dx
(1). f ( x ) f(x) f(x) 在 [ 0 , a ] [0,a] [0,a] ， [ 0 , a ] , ∫ 0 a f ( x ) d x = ∫ 0 a 2 f ( x ) d x + ∫ 0 a 2 f ( a − x ) d x \left[ {0,a} \right],\int_0^a {f\left( x \right)} dx = \int_0^{\frac{a}{2}} {f\left( x \right)} dx + \int_0^{\frac{a}{2}} {f\left( {a - x} \right)} dx [0,a],∫0af(x)dx=∫02af(x)dx+∫02af(a−x)dx

∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x \int_0^a {f\left( x \right)} dx = \int_0^a {f\left( {a - x} \right)} dx ∫0af(x)dx=∫0af(a−x)dx (这个公式用的比较多)

(2). f ( x ) f(x) f(x) 在 [ − a , a ] [-a,a] [−a,a] ， ∫ − a a f ( x ) d x = ∫ 0 a [ f ( x ) + f ( − x ) ] d x \int_{ - a}^a {f\left( x \right)} dx = \int_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]dx} ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx

补充： ∫ 0 π x f ( sin ⁡ x ) d x = π 2 ∫ 0 π f ( sin ⁡ x ) d x \int_{\rm{0}}^\pi {xf\left( {\sin x} \right)} dx = \frac{\pi }{2}\int_{\rm{0}}^\pi {f\left( {\sin x} \right)} dx ∫0πxf(sinx)dx=2π∫0πf(sinx)dx

2. 不定积分

例题1： ∫ 1 1 + x 4 d x \int {\frac{1}{{1 + {x^4}}}} dx ∫1+x41dx

= 1 2 ∫ 1 + x 2 1 + x 4 − x 2 − 1 1 + x 4 d x = 1 2 ∫ d ( x − 1 x ) ( x − 1 x ) 2 + 2 − 1 2 ∫ d ( x + 1 x ) ( x + 1 x ) 2 − 2 = 1 2 2 arctan ⁡ 1 2 ( x − 1 x ) − 1 4 2 ln ⁡ ∣ x + 1 x − 2 x + 1 x + 2 ∣ + c ( x ≠ 0 ) \begin{array}{l} = \frac{1}{2}\int {\frac{{1 + {x^2}}}{{1 + {x^4}}} - \frac{{{x^2} - 1}}{{1 + {x^4}}}} dx = \frac{1}{2}\int {\frac{{d\left( {x - \frac{1}{x}} \right)}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 2}}} - \frac{1}{2}\int {\frac{{d\left( {x + \frac{1}{x}} \right)}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 2}}} \\\;\\ = \frac{1}{{2\sqrt 2 }}\arctan \frac{1}{{\sqrt 2 }}\left( {x - \frac{1}{x}} \right) - \frac{1}{{4\sqrt 2 }}\ln \left| {\frac{{x + \frac{1}{x} - \sqrt 2 }}{{x + \frac{1}{x} + \sqrt 2 }}} \right| + c\left( {x \ne 0} \right)\end{array} =21∫1+x41+x2−1+x4x2−1dx=21∫(x−x1)2+2d(x−x1)−21∫(x+x1)2−2d(x+x1)=22 1arctan2 1(x−x1)−42 1ln∣∣∣x+x1+2 x+x1−2 ∣∣∣+c(x=0)

例题2： ∫ d x sin ⁡ 3 x cos ⁡ x \int {\frac{{dx}}{{{{\sin }^3}x\cos x}}} ∫sin3xcosxdx

= ∫ sin ⁡ 2 x + cos ⁡ 2 x sin ⁡ 3 x cos ⁡ x d x = ∫ d tan ⁡ x tan ⁡ x + ∫ d sin ⁡ x sin ⁡ 3 x = ln ⁡ ∣ tan ⁡ x ∣ − 1 2 sin ⁡ 2 x + C \begin{array}{l} = \int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^3}x\cos x}}dx} = \int {\frac{{d\tan x}}{{\tan x}}} + \int {\frac{{d\sin x}}{{{{\sin }^3}x}}} \\ \;\\= \ln \left| {\tan x} \right| - \frac{1}{{2{{\sin }^2}x}} + C\end{array} =∫sin3xcosxsin2x+cos2xdx=∫tanxdtanx+∫sin3xdsinx=ln∣tanx∣−2sin2x1+C

例题3： ∫ x + sin ⁡ x 1 + cos ⁡ x d x \int {\frac{{x + \sin x}}{{1 + \cos x}}} dx ∫1+cosxx+sinxdx

= ∫ x 1 + cos ⁡ x d x + ∫ sin ⁡ x 1 + cos ⁡ x d x = ∫ x cos ⁡ 2 x 2 d x 2 + ∫ sin ⁡ x 1 + cos ⁡ x d x = x tan ⁡ x 2 − ∫ tan ⁡ x 2 d x + ∫ tan ⁡ x 2 d x = x tan ⁡ x 2 + C \begin{array}{l} = \int {\frac{x}{{1 + \cos x}}} dx + \int {\frac{{\sin x}}{{1 + \cos x}}} dx = \int {\frac{x}{{{{\cos }^2}\frac{x}{2}}}} d\frac{x}{2} + \int {\frac{{\sin x}}{{1 + \cos x}}} dx\\\;\\= x\tan \frac{x}{2} - \int {\tan \frac{x}{2}} dx + \int {\tan \frac{x}{2}} dx = x\tan \frac{x}{2} + C \end{array} =∫1+cosxxdx+∫1+cosxsinxdx=∫cos22xxd2x+∫1+cosxsinxdx=xtan2x−∫tan2xdx+∫tan2xdx=xtan2x+C

例题4： ∫ 1 − x 1 + x d x x \int {\sqrt {\frac{{1 - x}}{{1 + x}}} } \frac{{dx}}{x} ∫1+x1−x xdx 令 t = 1 − x 1 + x , x = 1 − t 2 1 + t 2 , d x = − 4 t ( 1 + t 2 ) 2 d t {t = \sqrt {\frac{{1 - x}}{{1 + x}}} ,x = \frac{{1 - {t^2}}}{{1 + {t^2}}},dx = \frac{{ - 4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}dt} t=1+x1−x ,x=1+t21−t2,dx=(1+t2)2−4tdt

= ∫ t ⋅ − 4 t ( 1 + t 2 ) 2 1 − t 2 1 + t 2 d t = ∫ − 4 t 2 ( 1 + t 2 ) ( 1 − t 2 ) d t = 2 ∫ 1 1 + t 2 − 1 1 − t 2 d t {\rm{ = }}\int {t \cdot } \frac{{\frac{{ - 4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}}}{{\frac{{1 - {t^2}}}{{1 + {t^2}}}}}dt = \int {\frac{{ - 4{t^2}}}{{\left( {1 + {t^2}} \right)\left( {1 - {t^2}} \right)}}} dt = 2\int {\frac{1}{{1 + {t^2}}}} - \frac{1}{{1 - {t^2}}}dt =∫t⋅1+t21−t2(1+t2)2−4tdt=∫(1+t2)(1−t2)−4t2dt=2∫1+t21−1−t21dt

= 2 arctan ⁡ t + ln ⁡ ∣ t − 1 t + 1 ∣ = 2 arctan ⁡ 1 − x 1 + x + ln ⁡ ∣ 1 − x − 1 + x 1 − x + 1 + x ∣ + C = 2\arctan t + \ln \left| {\frac{{t - 1}}{{t + 1}}} \right| = 2\arctan \sqrt {\frac{{1 - x}}{{1 + x}}} + \ln \left| {\frac{{\sqrt {1 - x} - \sqrt {1 + x} }}{{\sqrt {1 - x} + \sqrt {1 + x} }}} \right| + C =2arctant+ln∣∣t+1t−1∣∣=2arctan1+x1−x +ln∣∣∣1−x +1+x 1−x −1+x ∣∣∣+C

例题5（较难）： ∫ e 2 x + 4 e x − 1 d x \int {\sqrt {{e^{2x}} + 4{e^x} - 1} } dx ∫e2x+4ex−1 dx 令 t = e x , d x = 1 t d t {t = {e^x},dx = \frac{1}{t}dt} t=ex,dx=t1dt 即为 ∫ t 2 + 4 t − 1 t d t \int {\frac{{\sqrt {{t^2} + 4t - 1} }}{t}dt} ∫tt2+4t−1 dt

解法一：

= t 2 + 4 t − 1 − ∫ t 2 + 2 t − t 2 − 4 t + 1 t 2 t 2 + 4 t − 1 d t = t 2 + 4 t − 1 + ∫ 2 t − 1 t 2 t 2 + 4 t − 1 d t = t 2 + 4 t − 1 + ∫ 2 t t 2 + 4 t − 1 d t − ∫ 1 t 2 t 2 + 4 t − 1 d t = t 2 + 4 t − 1 + 4 arctan ⁡ a − ( 4 arctan ⁡ a + 4 a − 2 1 + a 2 ) = t 2 + 4 t − 1 − 2 t + 10 t 2 + 2 t + t t 2 + 4 t − 1 + C \begin{array}{l} = \sqrt {{t^2} + 4t - 1} - \int {\frac{{{t^2} + 2t - {t^2} - 4t + 1}}{{{t^2}\sqrt {{t^2} + 4t - 1} }}} dt\\\;\\ = \sqrt {{t^2} + 4t - 1} + \int {\frac{{2t - 1}}{{{t^2}\sqrt {{t^2} + 4t - 1} }}} dt\\\;\\ = \sqrt {{t^2} + 4t - 1} + \int {\frac{2}{{t\sqrt {{t^2} + 4t - 1} }}} dt - \int {\frac{1}{{{t^2}\sqrt {{t^2} + 4t - 1} }}} dt = \sqrt {{t^2} + 4t - 1} + 4\arctan a - \left( {4\arctan a + \frac{{4a - 2}}{{1 + {a^2}}}} \right)\\\;\\ = \sqrt {{t^2} + 4t - 1} - \frac{2}{t} + \frac{{10}}{{{t^2} + 2t + t\sqrt {{t^2} + 4t - 1} }} + C \end{array} =t2+4t−1 −∫t2t2+4t−1 t2+2t−t2−4t+1dt=t2+4t−1 +∫t2t2+4t−1 2t−1dt=t2+4t−1 +∫tt2+4t−1 2dt−∫t2t2+4t−1 1dt=t2+4t−1 +4arctana−(4arctana+1+a24a−2)=t2+4t−1 −t2+t2+2t+tt2+4t−1 10+C

设 t 2 + 4 t − 1 = a − t → t = a 2 + 1 2 a + 4 , a − t = a 2 + 4 a − 1 2 a + 4 , d t = 2 a 2 + 8 a − 2 ( 2 a + 4 ) 2 d a \sqrt {{t^2} + 4t - 1} = a - t \to t = \frac{{{a^2} + 1}}{{2a + 4}},a - t = \frac{{{a^2} + 4a - 1}}{{2a + 4}},dt = \frac{{2{a^2} + 8a - 2}}{{{{\left( {2a + 4} \right)}^2}}}da t2+4t−1 =a−t→t=2a+4a2+1,a−t=2a+4a2+4a−1,dt=(2a+4)22a2+8a−2da （这个方法需要掌握）

∫ 2 t t 2 + 4 t − 1 d t = 2 ∫ 1 a 2 + 1 2 a + 4 ⋅ a 2 + 4 a − 1 2 a + 4 ⋅ 2 a 2 + 8 a − 2 ( 2 a + 4 ) 2 d a = 4 ∫ 1 a 2 + 1 d a = 4 arctan ⁡ a ∫ 1 t 2 t 2 + 4 t − 1 d t = ∫ 1 ( a 2 + 1 2 a + 4 ) 2 ⋅ a 2 + 4 a − 1 2 a + 4 ⋅ 2 a 2 + 8 a − 2 ( 2 a + 4 ) 2 d a = 4 ∫ a + 2 ( a 2 + 1 ) 2 d a = 4 ∫ a ( a 2 + 1 ) 2 d a + 8 ∫ 1 ( a 2 + 1 ) 2 d a = 4 arctan ⁡ a + 4 a 1 + a 2 − 2 a 2 + 1 = 4 arctan ⁡ a + 4 a − 2 1 + a 2 4 ∫ a ( a 2 + 1 ) 2 d a = 2 ∫ d ( a 2 + 1 ) ( a 2 + 1 ) 2 = − 2 a 2 + 1 8 ∫ 1 ( a 2 + 1 ) 2 d a 8 ∫ 1 sec ⁡ 4 θ ⋅ sec ⁡ 2 θ d θ = 2 ∫ 1 + cos ⁡ 2 θ d 2 θ = 4 θ + 2 sin ⁡ 2 θ = 4 arctan ⁡ a + 4 a 1 + a 2 \begin{array}{l} \int {\frac{2}{{t\sqrt {{t^2} + 4t - 1} }}} dt = 2\int {\frac{1}{{\frac{{{a^2} + 1}}{{2a + 4}} \cdot \frac{{{a^2} + 4a - 1}}{{2a + 4}}}}} \cdot \frac{{2{a^2} + 8a - 2}}{{{{\left( {2a + 4} \right)}^2}}}da = 4\int {\frac{1}{{{a^2} + 1}}} da = 4\arctan a\\\;\\ \int {\frac{1}{{{t^2}\sqrt {{t^2} + 4t - 1} }}} dt = \int {\frac{1}{{{{\left( {\frac{{{a^2} + 1}}{{2a + 4}}} \right)}^2} \cdot \frac{{{a^2} + 4a - 1}}{{2a + 4}}}}} \cdot \frac{{2{a^2} + 8a - 2}}{{{{\left( {2a + 4} \right)}^2}}}da = 4\int {\frac{{a + 2}}{{{{\left( {{a^2} + 1} \right)}^2}}}} da\\\;\\ = 4\int {\frac{a}{{{{\left( {{a^2} + 1} \right)}^2}}}} da + 8\int {\frac{1}{{{{\left( {{a^2} + 1} \right)}^2}}}} da = 4\arctan a + \frac{{4a}}{{1 + {a^2}}} - \frac{2}{{{a^2} + 1}} = 4\arctan a + \frac{{4a - 2}}{{1 + {a^2}}}\\\;\\ 4\int {\frac{a}{{{{\left( {{a^2} + 1} \right)}^2}}}} da = 2\int {\frac{{d\left( {{a^2} + 1} \right)}}{{{{\left( {{a^2} + 1} \right)}^2}}}} = - \frac{2}{{{a^2} + 1}}\\\;\\ 8\int {\frac{1}{{{{\left( {{a^2} + 1} \right)}^2}}}} da8\int {\frac{1}{{{{\sec }^4}\theta }}} \cdot {\sec ^2}\theta d\theta = 2\int {1 + \cos 2\theta } d2\theta = 4\theta + 2\sin 2\theta = 4\arctan a + \frac{{4a}}{{1 + {a^2}}} \end{array} ∫tt2+4t−1 2dt=2∫2a+4a2+1⋅2a+4a2+4a−11⋅(2a+4)22a2+8a−2da=4∫a2+11da=4arctana∫t2t2+4t−1 1dt=∫(2a+4a2+1)2⋅2a+4a2+4a−11⋅(2a+4)22a2+8a−2da=4∫(a2+1)2a+2da=4∫(a2+1)2ada+8∫(a2+1)21da=4arctana+1+a24a−a2+12=4arctana+1+a24a−24∫(a2+1)2ada=2∫(a2+1)2d(a2+1)=−a2+128∫(a2+1)21da8∫sec4θ1⋅sec2θdθ=2∫1+cos2θd2θ=4θ+2sin2θ=4arctana+1+a24a

解法二：(倒代换) 令 x = 1 t , d t = − 1 x 2 d x {x = \frac{1}{t},dt = - \frac{1}{{{x^2}}}dx} x=t1,dt=−x21dx

= ∫ x ⋅ 1 + 4 x − x 2 x 2 d 1 x = t 2 + 4 t − 1 + ∫ x − 2 x 1 + 4 x − x 2 d x = t 2 + 4 t − 1 + ∫ 1 1 + 4 x − x 2 d x + 2 ∫ − 1 x 1 + 4 x − x 2 d x = t 2 + 4 t − 1 + ∫ 1 1 + 4 x − x 2 d x + 2 ∫ − 1 x 2 1 x 2 + 4 x − 1 d x = t 2 + 4 t − 1 + 2 ln ⁡ ∣ t + 2 + t 2 + 4 t − 1 ∣ + arcsin ⁡ ( 1 − 2 t 5 t ) + C \begin{array}{l} {\rm{ = }}\int {x \cdot \sqrt {\frac{{1 + 4x - {x^2}}}{{{x^2}}}} } d\frac{1}{x} = \sqrt {{t^2} + 4t - 1} + \int {\frac{{x - 2}}{{x\sqrt {1 + 4x - {x^2}} }}} dx\\\;\\ = \sqrt {{t^2} + 4t - 1} + \int {\frac{1}{{\sqrt {1 + 4x - {x^2}} }}} dx + 2\int {\frac{{ - 1}}{{x\sqrt {1 + 4x - {x^2}} }}} dx\\\;\\ = \sqrt {{t^2} + 4t - 1} + \int {\frac{1}{{\sqrt {1 + 4x - {x^2}} }}} dx + 2\int {\frac{{\frac{{ - 1}}{{{x^2}}}}}{{\sqrt {\frac{1}{{{x^2}}} + \frac{4}{x} - 1} }}} dx\\\;\\ = \sqrt {{t^2} + 4t - 1} + 2\ln \left| {t + 2 + \sqrt {{t^2} + 4t - 1} } \right| + \arcsin \left( {\frac{{1 - 2t}}{{\sqrt 5 t}}} \right){\rm{ + }}C \end{array} =∫x⋅x21+4x−x2 dx1=t2+4t−1 +∫x1+4x−x2 x−2dx=t2+4t−1 +∫1+4x−x2 1dx+2∫x1+4x−x2 −1dx=t2+4t−1 +∫1+4x−x2 1dx+2∫x21+x4−1 x2−1dx=t2+4t−1 +2ln∣∣t+2+t2+4t−1 ∣∣+arcsin(5 t1−2t)+C

或 = t 2 + 4 t − 1 + 2 ln ⁡ ∣ t + 2 + t 2 + 4 t − 1 ∣ − arctan ⁡ ( 2 t − 1 t 2 + 4 t − 1 ) + C = \sqrt {{t^2} + 4t - 1} + 2\ln \left| {t + 2 + \sqrt {{t^2} + 4t - 1} } \right| - \arctan \left( {\frac{{2t - 1}}{{\sqrt {{t^2} + 4t - 1} }}} \right) + C =t2+4t−1 +2ln∣∣t+2+t2+4t−1 ∣∣−arctan(t2+4t−1 2t−1)+C

其中： ∫ 1 1 + 4 x − x 2 d x = ∫ 1 5 − ( x − 2 ) 2 d x = arcsin ⁡ ( x − 2 5 ) = arcsin ⁡ ( 1 − 2 t 5 t ) \int {\frac{1}{{\sqrt {1 + 4x - {x^2}} }}} dx = \int {\frac{1}{{\sqrt {5 - {{\left( {x - 2} \right)}^2}} }}dx} = \arcsin \left( {\frac{{x - 2}}{{\sqrt 5 }}} \right) = \arcsin \left( {\frac{{1 - 2t}}{{\sqrt 5 t}}} \right) ∫1+4x−x2 1dx=∫5−(x−2)2 1dx=arcsin(5 x−2)=arcsin(5 t1−2t)

∫ − 1 x 2 1 x 2 + 4 x − 1 d x = ∫ 1 ( 1 x + 2 ) 2 − 5 d 1 x = ∫ d t ( t + 2 ) 2 − 5 \begin{array}{l} \int {\frac{{\frac{{ - 1}}{{{x^2}}}}}{{\sqrt {\frac{1}{{{x^2}}} + \frac{4}{x} - 1} }}} dx = \int {\frac{1}{{\sqrt {{{\left( {\frac{1}{x} + 2} \right)}^2} - 5} }}} d\frac{1}{x}\\\;\\ = \int {\frac{{dt}}{{\sqrt {{{\left( {t + 2} \right)}^2} - 5} }}} \end{array} ∫x21+x4−1 x2−1dx=∫(x1+2)2−5 1dx1=∫(t+2)2−5 dt

令 t + 2 = 5 sec ⁡ x , d t = 5 sec ⁡ x tan ⁡ x d x {t + 2 = \sqrt 5 \sec x,dt = \sqrt 5 \sec x\tan xdx} t+2=5 secx,dt=5 secxtanxdx

∫ sec ⁡ x d x = l n ∣ sec ⁡ x + tan ⁡ x ∣ = ln ⁡ ∣ t + 2 + t 2 + 4 t − 1 ∣ \begin{array}{l} \int {\sec xdx} = ln\left| {\sec x + \tan x} \right| = \ln \left| {t + 2 + \sqrt {{t^2} + 4t - 1} } \right|\end{array} ∫secxdx=ln∣secx+tanx∣=ln∣∣t+2+t2+4t−1 ∣∣

练习题1： ∫ 1 + x + x 2 x \int {\frac{{\sqrt {{\rm{1 + }}x + {x^2}} }}{x}} ∫x1+x+x2

3. 定积分

3.1 积分定义解题

例1：（定义法求积分） ∫ a b 1 x d x \int_a^b {\frac{1}{x}} dx ∫abx1dx

解：令 x i = a ( b a ) i n , Δ x = x i + 1 − x i = a ( b a ) i n [ ( b a ) 1 n − 1 ] {x_i} = a{\left( {\frac{b}{a}} \right)^{\frac{i}{n}}},\Delta x = {x_{i + 1}} - {x_i} = a{\left( {\frac{b}{a}} \right)^{\frac{i}{n}}}\left[ {{{\left( {\frac{b}{a}} \right)}^{\frac{1}{n}}} - 1} \right] xi=a(ab)ni,Δx=xi+1−xi=a(ab)ni[(ab)n1−1]

= lim ⁡ n → ∞ ∑ i = 0 n 1 a ( b a ) i n ⋅ a ( b a ) i n [ ( b a ) 1 n − 1 ] = lim ⁡ n → ∞ n ⋅ [ ( b a ) 1 n − 1 ] = lim ⁡ n → ∞ [ ( b a ) 1 n − 1 ] 1 n ( a x − 1 ∼ x ln ⁡ a ) = lim ⁡ n → ∞ 1 n ln ⁡ ( b a ) 1 n = ln ⁡ ( b a ) \begin{array}{l} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 0}^n {\frac{1}{{a{{\left( {\frac{b}{a}} \right)}^{\frac{i}{n}}}}}} \cdot a{\left( {\frac{b}{a}} \right)^{\frac{i}{n}}}\left[ {{{\left( {\frac{b}{a}} \right)}^{\frac{1}{n}}} - 1} \right]\\\;\\ = \mathop {\lim }\limits_{n \to \infty } n \cdot \left[ {{{\left( {\frac{b}{a}} \right)}^{\frac{1}{n}}} - 1} \right] = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{{\left( {\frac{b}{a}} \right)}^{\frac{1}{n}}} - 1} \right]}}{{\frac{1}{n}}}\left( {{a^x} - 1 \sim x\ln a} \right)\\\;\\ = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{n}\ln \left( {\frac{b}{a}} \right)}}{{\frac{1}{n}}} = \ln \left( {\frac{b}{a}} \right) \end{array} =n→∞limi=0∑na(ab)ni1⋅a(ab)ni[(ab)n1−1]=n→∞limn⋅[(ab)n1−1]=n→∞limn1[(ab)n1−1](ax−1∼xlna)=n→∞limn1n1ln(ab)=ln(ab)

例2：（利用积分定义求极限）

I = lim ⁡ n → ∞ ( sin ⁡ π n n + 1 + sin ⁡ 2 π n n + 1 2 + ⋯ + sin ⁡ n π n n + 1 n ) = lim ⁡ n → ∞ ∑ i = 1 n sin ⁡ i π n n + 1 i ≠ f ( i n ) ⋅ 1 n sin ⁡ π i n n + 1 < sin ⁡ i π n n + 1 i < sin ⁡ π i n n ( i = 1 , 2 , ⋯ , n ) ∴ lim ⁡ n → ∞ ∑ i = 1 n sin ⁡ π i n ⋅ 1 n = ∫ 0 1 sin ⁡ π x d x = − 1 π cos ⁡ π x ∣ 0 1 = 2 π lim ⁡ n → ∞ n n + 1 ⋅ ∑ i = 1 n sin ⁡ π i n ⋅ 1 n = lim ⁡ n → ∞ n n + 1 ⋅ lim ⁡ n → ∞ ∑ i = 1 n sin ⁡ π i n ⋅ 1 n = 1 ⋅ ∫ 0 1 sin ⁡ π x d x = 2 π \begin{array}{l} {\rm{I}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\sin \frac{\pi }{n}}}{{n + 1}} + \frac{{\sin \frac{{2\pi }}{n}}}{{n + \frac{1}{2}}} + \cdots + \frac{{\sin \frac{{n\pi }}{n}}}{{n + \frac{1}{n}}}} \right)\\\;\\ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{\sin \frac{{i\pi }}{n}}}{{n + \frac{1}{i}}}} \ne f\left( {\frac{i}{n}} \right) \cdot \frac{1}{n}\\\;\\ \frac{{\sin \pi \frac{i}{n}}}{{n + 1}} < \frac{{\sin \frac{{i\pi }}{n}}}{{n + \frac{1}{i}}} < \frac{{\sin \pi \frac{i}{n}}}{n}\left( {i = 1,2, \cdots ,n} \right)\\\;\\ \therefore \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\sin \pi \frac{i}{n}} \cdot \frac{1}{n} = \int_0^1 {\sin \pi x} dx = - \frac{1}{\pi }\cos \pi x|_0^1 = \frac{2}{\pi }\\\;\\ \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}} \cdot \sum\limits_{i = 1}^n {\sin \pi \frac{i}{n}} \cdot \frac{1}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}} \cdot \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\sin \pi \frac{i}{n}} \cdot \frac{1}{n}\\\;\\ = 1 \cdot \int_0^1 {\sin \pi x} dx = \frac{2}{\pi } \end{array} I=n→∞lim(n+1sinnπ+n+21sinn2π+⋯+n+n1sinnnπ)=n→∞limi=1∑nn+i1sinniπ=f(ni)⋅n1n+1sinπni<n+i1sinniπ<nsinπni(i=1,2,⋯,n)∴n→∞limi=1∑nsinπni⋅n1=∫01sinπxdx=−π1cosπx∣01=π2n→∞limn+1n⋅i=1∑nsinπni⋅n1=n→∞limn+1n⋅n→∞limi=1∑nsinπni⋅n1=1⋅∫01sinπxdx=π2

有夹逼定理，得 I = lim ⁡ n → ∞ ∑ i = 1 n sin ⁡ i π n n + 1 i = 2 π {\rm{I}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{\sin \frac{{i\pi }}{n}}}{{n + \frac{1}{i}}}} = \frac{2}{\pi } I=n→∞limi=1∑nn+i1sinniπ=π2

练习题2：

( 1 ) . lim ⁡ n → ∞ ln ⁡ ( 1 + 1 n ) 2 ( 1 + 2 n ) 2 ⋯ ( 1 + n n ) 2 n ( 2 ) . lim ⁡ n → ∞ ( 1 n 2 + 1 + 1 n 2 + 2 2 + ⋯ + 1 n 2 + n 2 ) \begin{array}{l} \left( {\rm{1}} \right){\rm{.}}\mathop {\lim }\limits_{{\rm{n}} \to \infty } \ln \sqrt[n]{{{{\left( {1 + \frac{1}{n}} \right)}^2}{{\left( {1 + \frac{2}{n}} \right)}^2} \cdots {{\left( {1 + \frac{n}{n}} \right)}^2}}}\\ \left( 2 \right).\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\sqrt {{n^2} + 1} }} + \frac{1}{{\sqrt {{n^2} + {2^2}} }} + \cdots + \frac{1}{{\sqrt {{n^2} + {n^2}} }}} \right) \end{array} (1).n→∞limlnn(1+n1)2(1+n2)2⋯(1+nn)2 (2).n→∞lim(n2+1 1+n2+22 1+⋯+n2+n2 1)

3.2 积分技巧解题

例题1： ∫ 0 π 4 ln ⁡ ( 1 + t a n x ) d x \int_0^{\frac{\pi }{4}} {\ln (1 + tanx)} dx ∫04πln(1+tanx)dx

解：

= ∫ 0 π 4 ln ⁡ ( 1 + t a n ( π 4 − x ) ) d x = ∫ 0 π 4 ln ⁡ ( 2 1 + tan ⁡ x ) d x = ∫ 0 π 4 ln ⁡ 2 d x − ∫ 0 π 4 ln ⁡ ( 1 + t a n x ) d x → 原式 = 1 2 ∫ 0 π 4 ln ⁡ 2 d x = π 8 ln ⁡ 2 \begin{array}{l} = \int_0^{\frac{\pi }{4}} {\ln (1 + tan(\frac{\pi }{4} - x))} dx\\\;\\ = \int_0^{\frac{\pi }{4}} {\ln (\frac{2}{{1 + \tan x}})} dx\\\;\\ = \int_0^{\frac{\pi }{4}} {\ln 2dx} {\rm{ - }}\int_0^{\frac{\pi }{4}} {\ln (1 + tanx)} dx \to 原式 {\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_0^{\frac{\pi }{4}} {\ln 2dx} {\rm{ = }}\frac{\pi }{{\rm{8}}}\ln 2 \end{array} =∫04πln(1+tan(4π−x))dx=∫04πln(1+tanx2)dx=∫04πln2dx−∫04πln(1+tanx)dx→原式=21∫04πln2dx=8πln2

例题2： I = ∫ 2 4 x + 3 x + 3 + 9 − x d x I{\rm{ = }}\int_{\rm{2}}^{\rm{4}} {\frac{{\sqrt {x + 3} }}{{\sqrt {x + 3} + \sqrt {9 - x} }}} dx I=∫24x+3 +9−x x+3 dx

令 x + 3 = 9 − t , d x = − d t x + 3 = 9 - t,dx = - dt x+3=9−t,dx=−dt

I = ∫ 2 4 x + 3 x + 3 + 9 − x d x = − ∫ 4 2 9 − t 9 − t + 3 + t d t = ∫ 2 4 9 − x 9 − x + 3 + x d x = 1 2 [ ∫ 2 4 x + 3 x + 3 + 9 − x d x + ∫ 2 4 9 − x 9 − x + 3 + x d x ] = 1 2 ∫ 2 4 d x = 1 \begin{array}{l} I = \int_2^4 {\frac{{\sqrt {x + 3} }}{{\sqrt {x + 3} + \sqrt {9 - x} }}dx = - \int_4^2 {\frac{{\sqrt {9 - t} }}{{\sqrt {9 - t} + \sqrt {3 + t} }}} } dt = \int_2^4 {\frac{{\sqrt {9 - x} }}{{\sqrt {9 - x} + \sqrt {3 + x} }}} dx\\\;\\ = \frac{1}{2}\left[ {\int_2^4 {\frac{{\sqrt {x + 3} }}{{\sqrt {x + 3} + \sqrt {9 - x} }}dx + } \int_2^4 {\frac{{\sqrt {9 - x} }}{{\sqrt {9 - x} + \sqrt {3 + x} }}} dx} \right] = \frac{1}{2}\int_2^4 {dx} = 1 \end{array} I=∫24x+3 +9−x x+3 dx=−∫429−t +3+t 9−t dt=∫249−x +3+x 9−x dx=21[∫24x+3 +9−x x+3 dx+∫249−x +3+x 9−x dx]=21∫24dx=1

例题3： ∫ 0 a a − x x + a d x \int_{\rm{0}}^{\rm{a}} {\sqrt {\frac{{a - x}}{{x + a}}} } dx ∫0ax+aa−x dx

解：令 x = a cos ⁡ t , t ∈ ( 0 , π 2 ) , d x = − a sin ⁡ t d t {{\rm{x}} = a\cos t,t \in (0,\frac{\pi }{2})},{dx = - a\sin tdt} x=acost,t∈(0,2π),dx=−asintdt a − x x + a = 1 − cos ⁡ t 1 + cos ⁡ t = sin ⁡ 2 t 2 cos ⁡ 2 t 2 = tan ⁡ t 2 = sin ⁡ t 1 + cos ⁡ t = 1 − cos ⁡ t sin ⁡ t \sqrt {\frac{{a - x}}{{x + a}}} = \sqrt {\frac{{1 - \cos t}}{{1 + \cos t}}} = \sqrt {\frac{{{{\sin }^2}\frac{t}{2}}}{{{{\cos }^2}\frac{t}{2}}}} = \tan \frac{t}{2} = \frac{{\sin t}}{{1 + \cos t}} = \frac{{1 - \cos t}}{{\sin t}} x+aa−x =1+cost1−cost =cos22tsin22t =tan2t=1+costsint=sint1−cost (必须会)

= ∫ π 2 0 tan ⁡ t 2 ( − a s i n t ) d t = a ∫ 0 π 2 sin ⁡ 2 t 1 + cos ⁡ t d t = a ∫ 0 π 2 1 − cos ⁡ 2 t 1 + cos ⁡ t d t = a ∫ 0 π 2 1 + cos ⁡ t d t = π 2 a + a \begin{array}{l} = \int_{\frac{\pi }{2}}^0 {\tan \frac{t}{2}} ( - asint)dt\\\;\\ = a\int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}t}}{{1 + \cos t}}} dt\\\;\\ = a\int_0^{\frac{\pi }{2}} {\frac{{1 - {{\cos }^2}t}}{{1 + \cos t}}} dt\\\;\\ = a\int_0^{\frac{\pi }{2}} {1 + \cos t} dt\\\;\\ = \frac{\pi }{2}a + a \end{array} =∫2π0tan2t(−asint)dt=a∫02π1+costsin2tdt=a∫02π1+cost1−cos2tdt=a∫02π1+costdt=2πa+a

反常积分：

∫ 0 ∞ 1 1 + x 4 d x \int_{\rm{0}}^\infty {\frac{{\rm{1}}}{{{\rm{1 + }}{x^4}}}} dx ∫0∞1+x41dx 令 t = 1 x {t = \frac{1}{x}} t=x1

∫ 0 ∞ t 2 1 + t 4 d t = ∫ 0 ∞ x 2 1 + x 4 d x = 1 2 ∫ 0 ∞ 1 + x 2 1 + x 4 d x = 1 2 ∫ 0 ∞ d ( x − 1 x ) ( x − 1 x ) 2 + 2 = 1 2 2 arctan ⁡ 1 2 ( x − 1 x ) ∣ 0 ∞ = π 4 2 − ( − π 4 2 ) = π 2 2 \int_{\rm{0}}^\infty {\frac{{{t^2}}}{{{\rm{1 + }}{t^4}}}} dt = \int_{\rm{0}}^\infty {\frac{{{x^2}}}{{{\rm{1 + }}{x^4}}}} dx = \frac{1}{2}\int_0^\infty {\frac{{1 + {x^2}}}{{1 + {x^4}}}} dx\\\;\\ = \frac{1}{2}\int_0^\infty {\frac{{d\left( {x - \frac{1}{x}} \right)}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 2}}} = \left. {\frac{1}{{2\sqrt 2 }}\arctan \frac{1}{{\sqrt 2 }}\left( {x - \frac{1}{x}} \right)} \right|_0^\infty = \frac{\pi }{{4\sqrt 2 }} - \left( { - \frac{\pi }{{4\sqrt 2 }}} \right) = \frac{\pi }{{2\sqrt 2 }} ∫0∞1+t4t2dt=∫0∞1+x4x2dx=21∫0∞1+x41+x2dx=21∫0∞(x−x1)2+2d(x−x1)=22 1arctan2 1(x−x1)∣∣∣0∞=42 π−(−42 π)=22 π

（记忆）

I n = ∫ 0 π 2 sin ⁡ n x d x = ∫ 0 π 2 cos ⁡ n x d x = ∫ 0 π 2 sin ⁡ n − 1 x d ( − cos ⁡ x ) = [ − sin ⁡ n − 1 x cos ⁡ x ] 0 π 2 + ∫ 0 π 2 cos ⁡ x d ( sin ⁡ n − 1 x ) = ( n − 1 ) ∫ 0 π 2 sin ⁡ n − 2 x cos ⁡ 2 x d x = ( n − 1 ) ∫ 0 π 2 sin ⁡ n − 2 x ( 1 − sin ⁡ 2 x ) d x = ( n − 1 ) ∫ 0 π 2 sin ⁡ n − 2 x d x − ( n − 1 ) ∫ 0 π 2 sin ⁡ n x d x → I n = ( n − 1 ) I n − 2 − ( n − 1 ) I n \begin{array}{l} {I_n} = \int_0^{\frac{\pi }{2}} {{{\sin }^n}x} dx = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx = \int_0^{\frac{\pi }{2}} {{{\sin }^{n - 1}}x} d\left( { - \cos x} \right)\\\;\\ = \left[ { - {{\sin }^{n - 1}}x\cos x} \right]_0^{\frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\cos x} d\left( {{{\sin }^{n - 1}}x} \right)\\ = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\sin }^{n - 2}}x} {\cos ^2}xdx = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\sin }^{n - 2}}x} \left( {1 - {{\sin }^2}x} \right)dx\\\;\\ = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\sin }^{n - 2}}x} dx - \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\sin }^n}x} dx \to {I_n} = \left( {n - 1} \right){I_{n - 2}} - \left( {n - 1} \right){I_n}\end{array} In=∫02πsinnxdx=∫02πcosnxdx=∫02πsinn−1xd(−cosx)=[−sinn−1xcosx]02π+∫02πcosxd(sinn−1x)=(n−1)∫02πsinn−2xcos2xdx=(n−1)∫02πsinn−2x(1−sin2x)dx=(n−1)∫02πsinn−2xdx−(n−1)∫02πsinnxdx→In=(n−1)In−2−(n−1)In

I n = n − 1 n I n − 2 → n 为奇数 I 2 m + 1 = 2 m ( 2 m − 2 ) ⋯ 42 ( 2 m + 1 ) ( 2 m − 1 ) ⋯ 31 = ( 2 m ) ! ! ( 2 m + 1 ) ! ! {I_n} = \frac{{n - 1}}{n}{I_{n - 2}} \to n 为奇数{I_{2m + 1}} = \frac{{2m\left( {2m - 2} \right) \cdots 42}}{{\left( {2m + 1} \right)\left( {2m - 1} \right) \cdots 31}} = \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} In=nn−1In−2→n为奇数I2m+1=(2m+1)(2m−1)⋯312m(2m−2)⋯42=(2m+1)!!(2m)!!

n 为偶数 I 2 m = ( 2 m − 1 ) ( 2 m − 3 ) ⋯ 31 2 m ( 2 m − 2 ) ⋯ 42 = ( 2 m − 1 ) ! ! ( 2 m ) ! ! π 2 n为偶数{I_{2m}} = \frac{{\left( {2m - 1} \right)\left( {2m - 3} \right) \cdots 31}}{{2m\left( {2m - 2} \right) \cdots 42}} = \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} n为偶数I2m=2m(2m−2)⋯42(2m−1)(2m−3)⋯31=(2m)!!(2m−1)!!2π

练习题3：

（1） ∫ 0 a x 2 a − x a + x d x \int_0^a {{x^2}\sqrt {\frac{{a - x}}{{a + x}}} } dx ∫0ax2a+xa−x dx (2) ∫ 0 1 ln ⁡ ( 1 + x ) 1 + x 2 d x \int_0^1 {\frac{{\ln \left( {1 + x} \right)}}{{1 + {x^2}}}} dx ∫011+x2ln(1+x)dx

4. 练习解答

提示 题目解析都是由本人自己所写希望会有不同的做法与我分享，这里就不把答案直接放出来，本人会将答案放在另一篇blogs，等大家做完了点击这里一个链接：答案。

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微积分——巧解不定积分和定积分问题

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