[高等数学]换元积分法
一、知识点
(一) 第一类换元法
定理1
设 f ( u ) f(u) f(u) 具有原函数， u = φ ( x ) u=\varphi(x) u=φ(x) 可导，则有换元公式:
∫ f [ φ ( x ) ] φ ′ ( x ) d x = [ ∫ f ( u ) d u ] u = φ ( x ) . \int f[\varphi(x)]\varphi '(x)dx=[\int f(u)du]_{u=\varphi(x)}. ∫f[φ(x)]φ′(x)dx=[∫f(u)du]u=φ(x).
(二) 第二类换元法
定理2
设 x = φ ( x ) x=\varphi(x) x=φ(x) 是单调的、可导的函数，并且 φ ′ ( t ) ≠ 0 \varphi'(t)\neq 0 φ′(t)=0. 又设 f [ φ ( t ) ] φ ′ ( t ) f[\varphi(t)]\varphi'(t) f[φ(t)]φ′(t) 具有原函数，则有换元公式 ∫ f ( x ) d x = [ ∫ f [ φ ( t ) ] φ ′ ( t ) d t ] t = φ − 1 ( x ) ， \int f(x)dx=[\int f[\varphi(t)]\varphi'(t)dt]_{t=\varphi^{-1}(x)}， ∫f(x)dx=[∫f[φ(t)]φ′(t)dt]t=φ−1(x)， 其中， φ − 1 ( x ) \varphi^{-1}(x) φ−1(x) 是 x = φ ( t ) x=\varphi(t) x=φ(t) 的反函数.
二、练习题（求不定积分）
( 1 ) ∫ e 5 t d t = 1 5 ∫ e 5 t d ( 5 t ) = 1 5 e 5 t + C \begin{aligned} (1) &\int e^{5t}dt \\ & =\frac{1}{5}\int e^{5t}d(5t) \\ & =\frac{1}{5}e^{5t}+C \end{aligned} (1)∫e5tdt=51∫e5td(5t)=51e5t+C
( 2 ) ∫ ( 3 − 2 x ) 3 d x = − 1 2 ∫ ( 3 − 2 x ) 3 d ( 3 − 2 x ) = − 1 8 ( 3 − 2 x ) 4 + C \begin{aligned} (2) &\int(3-2x)^3dx\\ &=-\frac{1}{2}\int(3-2x)^3d(3-2x)\\ &=-\frac{1}{8}(3-2x)^4+C \end{aligned} (2)∫(3−2x)3dx=−21∫(3−2x)3d(3−2x)=−81(3−2x)4+C
( 3 ) ∫ d x 1 − 2 x = − 1 2 ∫ d ( 1 − 2 x ) 1 − 2 x = − 1 2 l n ∣ 1 − 2 x ∣ + C \begin{aligned} (3) &\int \frac{dx}{1-2x}\\ &=-\frac{1}{2}\int \frac{d(1-2x)}{1-2x}\\ &=-\frac{1}{2}ln|1-2x|+C \end{aligned} (3)∫1−2xdx=−21∫1−2xd(1−2x)=−21ln∣1−2x∣+C
( 4 ) ∫ d x 2 − 3 x 3 = − 1 3 ∫ d ( 2 − 3 x ) ( 2 − 3 x ) 1 3 = − 1 2 ( 2 − 3 x ) 2 3 + C \begin{aligned} (4) &\int\frac{dx}{\sqrt[3]{2-3x}}\\ &=-\frac{1}{3}\int\frac{d(2-3x)}{(2-3x)^{\frac{1}{3}}}\\ &=-\frac{1}{2}(2-3x)^{\frac{2}{3}}+C \end{aligned} (4)∫32−3xdx=−31∫(2−3x)31d(2−3x)=−21(2−3x)32+C
( 5 ) ∫ ( s i n a x − e x b ) d x = 1 a ∫ s i n a x d ( a x ) − b ∫ e x b d ( x b ) = − 1 a c o s a x − b e x b + C \begin{aligned} (5) &\int(sinax-e^{\frac{x}{b}})dx\\ &=\frac{1}{a}\int sinaxd(ax)-b\int e^{\frac{x}{b}}d(\frac{x}{b})\\ &=-\frac{1}{a}cosax-be^{\frac{x}{b}}+C \end{aligned} (5)∫(sinax−ebx)dx=a1∫sinaxd(ax)−b∫ebxd(bx)=−a1cosax−bebx+C
( 6 ) ∫ s i n t t d t = 2 ∫ s i n t d t = − 2 c o s t + C \begin{aligned} (6) &\int\frac{sin\sqrt{t}}{\sqrt{t}}dt\\ &=2\int sin\sqrt{t}d\sqrt{t}\\ &=-2cos\sqrt{t}+C \end{aligned} (6)∫tsintdt=2∫sintdt=−2cost+C
( 7 ) ∫ x e − x 2 d x = − 1 2 ∫ e − x 2 d ( − x 2 ) = − 1 2 e − x 2 + C \begin{aligned} (7) &\int xe^{-x^2}dx \\ &=-\frac{1}{2}\int e^{-x^2}d(-x^2) \\ &=-\frac{1}{2}e^{-x^2}+C \end{aligned} (7)∫xe−x2dx=−21∫e−x2d(−x2)=−21e−x2+C
( 8 ) ∫ x c o s x 2 d x = 1 2 ∫ c o s x 2 d ( x 2 ) = 1 2 s i n x 2 + C \begin{aligned} (8) &\int xcosx^2dx\\ &=\frac{1}{2}\int cosx^2d(x^2)\\ &=\frac{1}{2}sinx^2+C \end{aligned} (8)∫xcosx2dx=21∫cosx2d(x2)=21sinx2+C
( 9 ) ∫ x 2 − 3 x 2 d x = − 1 6 ∫ ( 2 − 3 x 2 ) − 1 2 d ( 2 − 3 x 2 ) = − 1 3 2 − 3 x 2 + C \begin{aligned} (9) &\int \frac{x}{\sqrt{2-3x^2}}dx\\ &=-\frac{1}{6}\int (2-3x^2)^{-\frac{1}{2}}d(2-3x^2)\\ &=-\frac{1}{3}\sqrt{2-3x^2}+C \end{aligned} (9)∫2−3x2xdx=−61∫(2−3x2)−21d(2−3x2)=−312−3x2+C
( 10 ) ∫ 3 x 3 1 − x 4 d x = − 3 4 ∫ d ( 1 − x 4 ) 1 − x 4 = − 3 4 l n ∣ 1 − x 4 ∣ + C \begin{aligned} (10) &\int\frac{3x^3}{1-x^4}dx\\ &=-\frac{3}{4}\int\frac{d(1-x^4)}{1-x^4}\\ &=-\frac{3}{4}ln|1-x^4|+C \end{aligned} (10)∫1−x43x3dx=−43∫1−x4d(1−x4)=−43ln∣1−x4∣+C
( 11 ) ∫ x + 1 x 2 + 2 x + 5 d x = 1 2 ∫ d ( x 2 + 2 x + 5 ) x 2 + 2 x + 5 = 1 2 l n ( x 2 + 2 x + 5 ) + C \begin{aligned} (11) &\int\frac{x+1}{x^2+2x+5}dx \\ &=\frac{1}{2}\int \frac{d(x^2+2x+5)}{x^2+2x+5}\\ &=\frac{1}{2}ln(x^2+2x+5)+C \end{aligned} (11)∫x2+2x+5x+1dx=21∫x2+2x+5d(x2+2x+5)=21ln(x2+2x+5)+C
( 12 ) ∫ c o s 2 ( ω t + φ ) s i n ( ω t + φ ) d t = − 1 ω ∫ c o s 2 ( ω t + φ ) d [ c o s ( ω t + φ ) ] = − 1 3 ω c o s 3 ( ω t + φ ) + C \begin{aligned} (12) &\int cos^2(\omega t+\varphi)sin(\omega t+\varphi)dt \\ &=-\frac{1}{\omega}\int cos^2(\omega t+\varphi)d[cos(\omega t+\varphi)]\\ &=-\frac{1}{3\omega}cos^3(\omega t+\varphi)+C \end{aligned} (12)∫cos2(ωt+φ)sin(ωt+φ)dt=−ω1∫cos2(ωt+φ)d[cos(ωt+φ)]=−3ω1cos3(ωt+φ)+C
( 13 ) ∫ s i n x c o s 3 x d x = − ∫ d ( c o s x ) c o s 3 x = 1 2 c o s − 2 x + C \begin{aligned} (13) &\int \frac{sinx}{cos^3x}dx\\ &=-\int \frac{d(cosx)}{cos^3x}\\ &=\frac{1}{2}cos^{-2}x+C \end{aligned} (13)∫cos3xsinxdx=−∫cos3xd(cosx)=21cos−2x+C
( 14 ) ∫ s i n x + c o s x s i n x − c o s x 3 d x = ∫ ( s i n x − c o s x ) − 1 3 d ( s i n x − c o s x ) = 3 2 ( s i n x − c o s x ) 2 3 + C \begin{aligned} (14) &\int\frac{sinx+cosx}{\sqrt[3]{sinx-cosx}}dx \\ &=\int(sinx-cosx)^{-\frac{1}{3}}d(sinx-cosx)\\ &=\frac{3}{2}(sinx-cosx)^{\frac{2}{3}}+C \end{aligned} (14)∫3sinx−cosxsinx+cosxdx=∫(sinx−cosx)−31d(sinx−cosx)=23(sinx−cosx)32+C
( 15 ) ∫ t a n 10 x s e c 2 x d x = ∫ t a n 10 d ( t a n x ) = 1 11 t a n 11 x + C \begin{aligned} (15) &\int tan^{10}xsec^2xdx \\ &=\int tan^{10}d(tanx)\\ &=\frac{1}{11}tan^{11}x+C \end{aligned} (15)∫tan10xsec2xdx=∫tan10d(tanx)=111tan11x+C
( 16 ) ∫ d x x ( l n x ) l n ( l n x ) d x = ∫ d ( l n x ) ( l n x ) l n ( l n x ) = ∫ d [ l n ( l n x ) ] l n ( l n x ) = l n ∣ l n ( l n x ) ∣ + C ( x > 1 ) \begin{aligned} (16) &\int \frac{dx}{x(lnx)ln(lnx)}dx\\ &=\int \frac{d(lnx)}{(lnx)ln(lnx)}\\ &=\int \frac{d[ln(lnx)]}{ln(lnx)}\\ &=ln|ln(lnx)|+C(x>1) \end{aligned} (16)∫x(lnx)ln(lnx)dxdx=∫(lnx)ln(lnx)d(lnx)=∫ln(lnx)d[ln(lnx)]=ln∣ln(lnx)∣+C(x>1)
( 17 ) ∫ d x ( a r c s i n x ) 2 1 − x 2 = ∫ d ( a r c s i n x ) ( a r c s i n x ) 2 = − 1 a r c s i n x + C \begin{aligned} (17) &\int\frac{dx}{(arcsinx)^2\sqrt{1-x^2}}\\ &=\int \frac{d(arcsinx)}{(arcsinx)^2}\\ &=-\frac{1}{arcsinx}+C \end{aligned} (17)∫(arcsinx)21−x2dx=∫(arcsinx)2d(arcsinx)=−arcsinx1+C
( 18 ) ∫ 1 0 2 a r c c o s x 1 − x 2 = − 1 2 ∫ 1 0 2 a r c c o s x d ( 2 a r c c o s x ) = − 1 0 2 a r c c o s x 2 l n 10 + C \begin{aligned} (18) &\int \frac{10^{2arccosx}}{\sqrt{1-x^2}}\\ &=-\frac{1}{2}\int 10^{2arccosx}d(2arccosx)\\ &=-\frac{10^{2arccosx}}{2ln10}+C \end{aligned} (18)∫1−x2102arccosx=−21∫102arccosxd(2arccosx)=−2ln10102arccosx+C
( 19 ) ∫ t a n 1 + x 2 ⋅ x d x 1 + x 2 = ∫ t a n 1 + x 2 d ( 1 + x 2 ) = − ∫ d ( c o s 1 + x 2 ) c o s 1 + x 2 = − l n ∣ c o s 1 + x 2 ∣ + C \begin{aligned} (19) &\int tan\sqrt{1+x^2}\cdot \frac{xdx}{\sqrt{1+x^2}}\\ &=\int tan\sqrt{1+x^2}d(\sqrt{1+x^2})\\ &=-\int \frac{d(cos\sqrt{1+x^2})}{cos\sqrt{1+x^2}}\\ &=-ln|cos\sqrt{1+x^2}|+C \end{aligned} (19)∫tan1+x2⋅1+x2xdx=∫tan1+x2d(1+x2)=−∫cos1+x2d(cos1+x2)=−ln∣cos1+x2∣+C
( 20 ) ∫ a r c t a n x x ( 1 + x ) d x = 2 ∫ a r c t a n x 1 + ( x ) 2 d x = 2 ∫ a r c t a n x d ( a r c t a n x ) = ( a r c t a n x ) 2 + C \begin{aligned} (20) &\int \frac{arctan\sqrt{x}}{\sqrt{x}(1+x)}dx\\ &=2\int \frac{arctan\sqrt{x}}{1+(\sqrt{x})^2}d\sqrt{x}\\ &=2\int arctan\sqrt{x}d(arctan\sqrt{x})\\ &=(arctan\sqrt{x})^2+C \end{aligned} (20)∫x(1+x)arctanxdx=2∫1+(x)2arctanxdx=2∫arctanxd(arctanx)=(arctanx)2+C
( 21 ) ∫ 1 + l n x ( x l n x ) 2 d x = ∫ d ( x l n x ) ( x l n x ) 2 = − 1 x l n x + C \begin{aligned} (21) &\int \frac{1+lnx}{(xlnx)^2}dx\\ &=\int \frac{d(xlnx)}{(xlnx)^2}\\ &=-\frac{1}{xlnx}+C \end{aligned} (21)∫(xlnx)21+lnxdx=∫(xlnx)2d(xlnx)=−xlnx1+C
( 22 ) ∫ d x s i n x c o s x = ∫ s e c 2 x t a n x d x = ∫ d ( t a n x ) t a n x = l n ∣ t a n x ∣ + C \begin{aligned} (22) &\int \frac{dx}{sinxcosx}\\ &=\int\frac{sec^2x}{tanx}dx\\ &=\int \frac{d(tanx)}{tanx}\\ &=ln|tanx|+C \end{aligned} (22)∫sinxcosxdx=∫tanxsec2xdx=∫tanxd(tanx)=ln∣tanx∣+C
( 23 ) ∫ l n t a n x c o s x s i n x d x = ∫ l n t a n x c o s 2 x t a n x d x = ∫ l n t a n x t a n x d ( t a n x ) = ∫ l n t a n x d ( l n t a n x ) = 1 2 ( l n t a n x ) 2 + C \begin{aligned} (23) &\int \frac{lntanx}{cosxsinx}dx\\ &=\int\frac{lntanx}{cos^2xtanx}dx\\ &=\int\frac{lntanx}{tanx}d(tanx)\\ &=\int lntanxd(lntanx)\\ &=\frac{1}{2}(lntanx)^2+C \end{aligned} (23)∫cosxsinxlntanxdx=∫cos2xtanxlntanxdx=∫tanxlntanxd(tanx)=∫lntanxd(lntanx)=21(lntanx)2+C
( 24 ) ∫ c o s 3 x d x = ∫ ( 1 − s i n 2 x ) d ( s i n x ) = s i n x − 1 3 s i n 3 x + C \begin{aligned} (24) &\int cos^3xdx\\ &=\int (1-sin^2x)d(sinx)\\ &=sinx-\frac{1}{3}sin^3x+C \end{aligned} (24)∫cos3xdx=∫(1−sin2x)d(sinx)=sinx−31sin3x+C
( 25 ) ∫ c o s 2 ( ω t + φ ) d t = ∫ 1 + c o s 2 ( ω t + φ ) 2 d t = ∫ d t 2 + 1 4 ω ∫ c o s 2 ( ω t + φ ) d [ 2 ( ω t + φ ) ] = 1 2 t + 1 4 ω s i n 2 ( ω t + φ ) + C \begin{aligned} (25) &\int cos^2(\omega t+\varphi)dt\\ &=\int \frac{1+cos2(\omega t+\varphi)}{2}dt\\ &=\int \frac{dt}{2}+\frac{1}{4\omega }\int cos2(\omega t+\varphi)d[2(\omega t+\varphi)]\\ &=\frac{1}{2}t+\frac{1}{4\omega}sin2(\omega t+\varphi)+C \end{aligned} (25)∫cos2(ωt+φ)dt=∫21+cos2(ωt+φ)dt=∫2dt+4ω1∫cos2(ωt+φ)d[2(ωt+φ)]=21t+4ω1sin2(ωt+φ)+C
( 26 ) ∫ s i n 2 x c o s 3 x d x = ∫ 1 2 ( s i n 5 x − s i n x ) d x = 1 2 c o s x − 1 10 c o s 5 x + C \begin{aligned} (26) &\int sin2xcos3xdx \\ &=\int \frac{1}{2}(sin5x-sinx)dx\\ &=\frac{1}{2}cosx-\frac{1}{10}cos5x+C \end{aligned} (26)∫sin2xcos3xdx=∫21(sin5x−sinx)dx=21cosx−101cos5x+C
( 27 ) ∫ c o s x c o s x 2 d x = ∫ 1 2 ( c o s 3 2 x + c o s x 2 ) d x = 1 2 × 2 3 ∫ c o s 3 2 x d ( 3 2 x ) + 1 2 × 2 ∫ c o s x 2 d ( x 2 ) = 1 3 s i n 3 2 x + s i n x 2 + C \begin{aligned} (27) &\int cosxcos\frac{x}{2}dx\\ &=\int \frac{1}{2}(cos\frac{3}{2}x+cos\frac{x}{2})dx\\ &=\frac{1}{2}\times \frac{2}{3}\int cos\frac{3}{2}xd(\frac{3}{2}x)+\frac{1}{2}\times2\int cos\frac{x}{2}d(\frac{x}{2})\\ &=\frac{1}{3}sin\frac{3}{2}x+sin\frac{x}{2}+C \end{aligned} (27)∫cosxcos2xdx=∫21(cos23x+cos2x)dx=21×32∫cos23xd(23x)+21×2∫cos2xd(2x)=31sin23x+sin2x+C
( 28 ) ∫ s i n 5 x s i n 7 x d x = ∫ − 1 2 ( c o s 12 x − c o s 2 x ) d x = − 1 2 ∫ c o s 12 x d x + 1 2 ∫ c o s 2 x d x = 1 4 s i n 2 x − 1 24 s i n 12 x + C \begin{aligned} (28) &\int sin5xsin7xdx\\ &=\int -\frac{1}{2}(cos12x-cos2x)dx\\ &=-\frac{1}{2}\int cos12xdx+\frac{1}{2}\int cos2xdx\\ &=\frac{1}{4}sin2x-\frac{1}{24}sin12x+C \end{aligned} (28)∫sin5xsin7xdx=∫−21(cos12x−cos2x)dx=−21∫cos12xdx+21∫cos2xdx=41sin2x−241sin12x+C
( 29 ) ∫ t a n 3 x s e c x d x = ∫ t a n 2 x d ( s e c x ) = ∫ ( s e c 2 x − 1 ) d ( s e c x ) = 1 3 s e c 3 x − s e c x + C \begin{aligned} (29) &\int tan^3xsecxdx\\ &=\int tan^2xd(secx)\\ &=\int (sec^2x-1)d(secx)\\ &=\frac{1}{3}sec^3x-secx+C \end{aligned} (29)∫tan3xsecxdx=∫tan2xd(secx)=∫(sec2x−1)d(secx)=31sec3x−secx+C
( 30 ) ∫ d x e x + e − x = ∫ e x d x e 2 x + 1 = ∫ d ( e x ) 1 + ( e x ) 2 = a r c t a n ( e x ) + C \begin{aligned} (30) &\int \frac{dx}{e^x+e^{-x}}\\ &=\int \frac{e^xdx}{e^{2x}+1}\\ &=\int \frac{d(e^x)}{1+(e^x)^2}\\ &=arctan(e^x)+C \end{aligned} (30)∫ex+e−xdx=∫e2x+1exdx=∫1+(ex)2d(ex)=arctan(ex)+C
( 31 ) ∫ 1 − x 9 − 4 x 2 d x = 1 2 ∫ d ( 2 x ) 3 2 − ( 2 x ) 2 + 1 8 ∫ ( 9 − 4 x 2 ) − 1 2 d ( 9 − 4 x 2 ) = 1 2 a r c s i n 2 x 3 + 1 4 9 − 4 x 2 + C \begin{aligned} (31) &\int \frac{1-x}{\sqrt{9-4x^2}}dx\\ &=\frac{1}{2}\int \frac{d(2x)}{\sqrt{3^2-(2x)^2}}+\frac{1}{8}\int (9-4x^2)^{-\frac{1}{2}}d(9-4x^2)\\ &=\frac{1}{2}arcsin\frac{2x}{3}+\frac{1}{4}\sqrt{9-4x^2}+C \end{aligned} (31)∫9−4x21−xdx=21∫32−(2x)2d(2x)+81∫(9−4x2)−21d(9−4x2)=21arcsin32x+419−4x2+C
( 32 ) ∫ x 3 9 + x 2 d x = 1 2 ∫ x 2 9 + x 2 d ( x 2 ) = 1 2 ∫ x 2 + 9 − 9 x 2 + 9 d ( x 2 ) = 1 2 ∫ d ( x 2 ) − 9 2 ∫ d ( x 2 + 9 ) x 2 + 9 = x 2 2 − 9 2 l n ( x 2 + 9 ) + C \begin{aligned} (32) &\int \frac{x^3}{9+x^2}dx\\ &=\frac{1}{2}\int \frac{x^2}{9+x^2}d(x^2)\\ &=\frac{1}{2}\int \frac{x^2+9-9}{x^2+9}d(x^2)\\ &=\frac{1}{2}\int d(x^2)-\frac{9}{2}\int \frac{d(x^2+9)}{x^2+9}\\ &=\frac{x^2}{2}-\frac{9}{2}ln(x^2+9)+C \end{aligned} (32)∫9+x2x3dx=21∫9+x2x2d(x2)=21∫x2+9x2+9−9d(x2)=21∫d(x2)−29∫x2+9d(x2+9)=2x2−29ln(x2+9)+C
( 33 ) ∫ d x 2 x 2 − 1 = ∫ d x ( 2 x − 1 ) ( 2 x + 1 ) = 1 2 2 ∫ ( 1 2 x − 1 − 1 2 x + 1 ) d 2 x = 1 2 2 ( l n ∣ 2 x − 1 ∣ − l n ∣ 2 x + 1 ∣ ) + C = 1 2 2 l n ∣ 2 x − 1 2 x + 1 ∣ + C \begin{aligned} (33) &\int \frac{dx}{2x^2-1}\\ &=\int \frac{dx}{(\sqrt{2}x-1)(\sqrt{2}x+1)}\\ &=\frac{1}{2\sqrt{2}}\int (\frac{1}{\sqrt{2}x-1}-\frac{1}{\sqrt{2}x+1})d\sqrt{2}x\\ &=\frac{1}{2\sqrt{2}}(ln|\sqrt{2}x-1|-ln|\sqrt{2}x+1|)+C\\ &=\frac{1}{2\sqrt{2}}ln\begin{vmatrix}\frac{\sqrt{2}x-1}{\sqrt{2}x+1}\end{vmatrix}+C \end{aligned} (33)∫2x2−1dx=∫(2x−1)(2x+1)dx=221∫(2x−11−2x+11)d2x=221(ln∣2x−1∣−ln∣2x+1∣)+C=221ln 2x+12x−1 +C
( 34 ) ∫ d x ( x + 1 ) ( x − 2 ) = 1 3 ∫ ( 1 x − 2 − 1 x + 1 ) d x = 1 3 l n ∣ x − 2 x + 1 ∣ + C \begin{aligned} (34) &\int \frac{dx}{(x+1)(x-2)}\\ &=\frac{1}{3}\int (\frac{1}{x-2}-\frac{1}{x+1})dx\\ &=\frac{1}{3}ln\begin{vmatrix}\frac{x-2}{x+1}\end{vmatrix}+C \end{aligned} (34)∫(x+1)(x−2)dx=31∫(x−21−x+11)dx=31ln x+1x−2 +C
( 35 ) ∫ x x 2 − x − 2 d x = ∫ x ( x − 2 ) ( x + 1 ) d x = 1 3 ∫ ( 2 x − 2 + 1 x + 1 ) d x = 2 3 l n ∣ x − 2 ∣ + 1 3 l n ∣ x + 1 ∣ + C \begin{aligned} (35) &\int \frac{x}{x^2-x-2}dx\\ &=\int \frac{x}{(x-2)(x+1)}dx\\ &=\frac{1}{3}\int(\frac{2}{x-2}+\frac{1}{x+1})dx\\ &=\frac{2}{3}ln|x-2|+\frac{1}{3}ln|x+1|+C \end{aligned} (35)∫x2−x−2xdx=∫(x−2)(x+1)xdx=31∫(x−22+x+11)dx=32ln∣x−2∣+31ln∣x+1∣+C
( 36 ) ∫ x 2 d x a 2 − x 2 ( a > 0 ) = ∫ a 2 ( x a ) 2 a 1 − ( x a ) 2 a d ( x a ) = a 2 ∫ ( x a ) 2 1 − ( x a ) 2 d ( x a ) ( 令 t = a r c s i n ( x a ) ) = a 2 ∫ s i n 2 t d ( t ) = a 2 ∫ 1 − c o s 2 t 2 d t = a 2 t 2 − a 2 s i n 2 t 4 + C = a 2 a r c s i n x a 2 − x a 2 − x 2 a 2 + C \begin{aligned} (36) &\int \frac{x^2dx}{\sqrt{a^2-x^2}}(a>0)\\ &=\int \frac{a^2(\frac{x}{a})^2}{a\sqrt{1-(\frac{x}{a})^2}}ad(\frac{x}{a})\\ &=a^2\int\frac{(\frac{x}{a})^2}{\sqrt{1-(\frac{x}{a})^2}}d(\frac{x}{a})\\ &(令t=arcsin(\frac{x}{a}))\\ &=a^2\int sin^2td(t)\\ &=a^2\int \frac{1-cos2t}{2}dt\\ &=\frac{a^2t}{2}-\frac{a^2sin2t}{4}+C\\ &=\frac{a^2arcsin\frac{x}{a}}{2}-\frac{x\sqrt{a^2-x^2}}{a^2}+C \end{aligned} (36)∫a2−x2x2dx(a>0)=∫a1−(ax)2a2(ax)2ad(ax)=a2∫1−(ax)2(ax)2d(ax)(令t=arcsin(ax))=a2∫sin2td(t)=a2∫21−cos2tdt=2a2t−4a2sin2t+C=2a2arcsinax−a2xa2−x2+C
( 37 ) ∫ d x x x 2 − 1 \begin{aligned} (37) &\int \frac{dx}{x\sqrt{x^2-1}}\end{aligned} (37)∫xx2−1dx
当 x > 1 x>1 x>1 时：
∫ d x x x 2 − 1 = ∫ d x x 2 1 − ( 1 x ) 2 = ∫ − d ( 1 x ) 1 − ( 1 x ) 2 = a r c c o s 1 x + C \begin{aligned} &\int \frac{dx}{x\sqrt{x^2-1}}\\ &=\int \frac{dx}{x^2\sqrt{1-(\frac{1}{x})^2}}\\ &=\int \frac{-d(\frac{1}{x})}{\sqrt{1-(\frac{1}{x})^2}}\\ &=arccos\frac{1}{x}+C \end{aligned} ∫xx2−1dx=∫x21−(x1)2dx=∫1−(x1)2−d(x1)=arccosx1+C
当 x < − 1 x<-1 x<−1 时：
∫ d x x x 2 − 1 = ∫ − d x x 2 1 − ( 1 x ) 2 = ∫ − d ( − 1 x ) 1 − ( − 1 x ) 2 = a r c c o s − 1 x + C \begin{aligned} &\int \frac{dx}{x\sqrt{x^2-1}}\\ &=\int \frac{-dx}{x^2\sqrt{1-(\frac{1}{x})^2}}\\ &=\int \frac{-d(-\frac{1}{x})}{\sqrt{1-(-\frac{1}{x})^2}}\\ &=arccos\frac{-1}{x}+C \end{aligned} ∫xx2−1dx=∫x21−(x1)2−dx=∫1−(−x1)2−d(−x1)=arccosx−1+C
学习资料：《高等数学（第六版）》 ，同济大学数学系 编
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