第2节 换元积分法
一、基础题
1 求下列不定积分
(1) $\displaystyle \int (3x+2)^{10}\,dx$；
解：
令 $u=3x+2$，则 $du=3dx$，$dx=\dfrac{1}{3}du$，故
$$\int (3x+2)^{10}\,dx=\int u^{10}\cdot \frac{1}{3}du =\frac{1}{3}\cdot \frac{u^{11}}{11} =\frac{(3x+2)^{11}}{33}+C.$$(2) $\displaystyle \int \frac{x+1}{x^{2}+2x+2}\,dx$；
解：
$x^{2}+2x+2=(x+1)^{2}+1$，令 $u=x+1$，则 $du=dx$，
$$\int \frac{x+1}{x^{2}+2x+2}\,dx =\int \frac{u}{u^{2}+1}du =\frac{1}{2}\ln(u^{2}+1)+C =\frac{1}{2}\ln(x^{2}+2x+2)+C.$$(3) $\displaystyle \int \frac{1}{1+2x^{2}}\,dx$；
解：
写成 $\displaystyle \int \frac{1}{1+(\sqrt{2}x)^{2}}\,dx$，令 $u=\sqrt{2}x$，则 $du=\sqrt{2}dx$，
$$\int \frac{1}{1+2x^{2}}\,dx =\frac{1}{\sqrt{2}}\int \frac{1}{1+u^{2}}\,du =\frac{1}{\sqrt{2}}\arctan u +C =\frac{1}{\sqrt{2}}\arctan(\sqrt{2}x)+C.$$(4) $\displaystyle \int \frac{1}{\sqrt[3]{2-3x}}\,dx$；
解：
写成 $\displaystyle \int (2-3x)^{-\tfrac{1}{3}}dx$。令 $u=2-3x$，则 $du=-3dx$，$dx=-\dfrac{1}{3}du$，
$$\int (2-3x)^{-\tfrac{1}{3}}dx =-\frac{1}{3}\int u^{-\tfrac{1}{3}}du =-\frac{1}{3}\cdot \frac{u^{\tfrac{2}{3}}}{\tfrac{2}{3}} =-\frac{1}{2}(2-3x)^{\tfrac{2}{3}}+C.$$(5) $\displaystyle \int \frac{3x^{5}}{1-x^{6}}\,dx$；
解：
令 $u=1-x^{6}$，则 $du=-6x^{5}dx$，有 $3x^{5}dx=-\dfrac{1}{2}du$，
$$\int \frac{3x^{5}}{1-x^{6}}\,dx =-\frac{1}{2}\int \frac{1}{u}du =-\frac{1}{2}\ln|u|+C =-\frac{1}{2}\ln|1-x^{6}|+C.$$(6) $\displaystyle \int \frac{4x}{\sqrt{2-3x^{2}}}\,dx$；
解：
令 $u=2-3x^{2}$，则 $du=-6x\,dx$，$4x\,dx=-\dfrac{2}{3}du$，
$$\int \frac{4x}{\sqrt{2-3x^{2}}}\,dx =-\frac{2}{3}\int u^{-\tfrac{1}{2}}du =-\frac{2}{3}\cdot 2u^{\tfrac{1}{2}}+C =-\frac{4}{3}\sqrt{2-3x^{2}}+C.$$(7) $\displaystyle \int 3\cos^{2}3x\sin3x\,dx$；
解：
令 $u=\cos3x$，则 $du=-3\sin3x\,dx$，$-du=3\sin3x\,dx$，
$$\int 3\cos^{2}3x\sin3x\,dx =\int u^{2}(-du) =-\int u^{2}du =-\frac{u^{3}}{3}+C =-\frac{\cos^{3}3x}{3}+C.$$(8) $\displaystyle \int \frac{\sin \dfrac{1}{x}}{x^{2}}\,dx$；
解：
令 $u=\dfrac{1}{x}$，则 $du=-\dfrac{1}{x^{2}}dx$，故
$$\int \frac{\sin\dfrac{1}{x}}{x^{2}}\,dx =-\int \sin u\,du =\cos u +C =\cos\frac{1}{x}+C.$$(9) $\displaystyle \int 8\tan^{6}x\sec^{2}x\,dx$；
解：
令 $u=\tan x$，则 $du=\sec^{2}x\,dx$，
$$\int 8\tan^{6}x\sec^{2}x\,dx =8\int u^{6}du =8\cdot \frac{u^{7}}{7}+C =\frac{8}{7}\tan^{7}x+C.$$(10) $\displaystyle \int \frac{\sin\sqrt{t}}{2\sqrt{t}}\,dt$；
解：
令 $u=\sqrt{t}$，则 $t=u^{2}$，$dt=2u\,du$，且 $\dfrac{1}{2\sqrt{t}}dt=\dfrac{1}{2u}\cdot 2u\,du=du$，
$$\int \frac{\sin\sqrt{t}}{2\sqrt{t}}\,dt =\int \sin u\,du =-\cos u +C =-\cos\sqrt{t}+C.$$(11) $\displaystyle \int \frac{\sqrt{1+\ln x}}{3x}\,dx$；
解：
令 $u=1+\ln x$，则 $du=\dfrac{1}{x}dx$，
$$\int \frac{\sqrt{1+\ln x}}{3x}\,dx =\frac{1}{3}\int u^{\tfrac{1}{2}}du =\frac{1}{3}\cdot \frac{2}{3}u^{\tfrac{3}{2}}+C =\frac{2}{9}(1+\ln x)^{\tfrac{3}{2}}+C.$$(12) $\displaystyle \int \frac{(\arctan x)^{2}}{1+x^{2}}\,dx$；
解：
令 $u=\arctan x$，则 $du=\dfrac{1}{1+x^{2}}dx$，
$$\int \frac{(\arctan x)^{2}}{1+x^{2}}\,dx =\int u^{2}du =\frac{u^{3}}{3}+C =\frac{(\arctan x)^{3}}{3}+C.$$2 求下列不定积分：
(1) $\displaystyle \int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}}\,dx(a>0)$；
解：
写 $x^{2}=a^{2}-(a^{2}-x^{2})$，则
$$\int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}}dx =a^{2}\int \frac{1}{\sqrt{a^{2}-x^{2}}}dx -\int \sqrt{a^{2}-x^{2}}\,dx.$$由已知公式
$$\int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=\arcsin\frac{x}{a}+C,$$$$\int \sqrt{a^{2}-x^{2}}\,dx =\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\arcsin\frac{x}{a}+C,$$代入得
$$\int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}}dx =a^{2}\arcsin\frac{x}{a} -\left(\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\arcsin\frac{x}{a}\right)+C$$$$=\frac{a^{2}}{2}\arcsin\frac{x}{a}-\frac{x}{2}\sqrt{a^{2}-x^{2}}+C.$$(2) $\displaystyle \int \frac{1}{x\sqrt{x^{2}-1}}\,dx$；
解：
注意到对 $x>1$ 有
$$\frac{d}{dx}\arccos\frac{1}{x} =\frac{1}{x\sqrt{x^{2}-1}}.$$因此
$$\int \frac{1}{x\sqrt{x^{2}-1}}\,dx =\arccos\frac{1}{x}+C.$$(3) $\displaystyle \int \frac{\sqrt{x^{2}-4}}{x}\,dx$；
解：
对 $x>2$，令 $x=2\sec\theta$，则
$$\sqrt{x^{2}-4}=\sqrt{4\sec^{2}\theta-4}=2\tan\theta,\quad dx=2\sec\theta\tan\theta\,d\theta.$$于是
$$\int \frac{\sqrt{x^{2}-4}}{x}\,dx =\int \frac{2\tan\theta}{2\sec\theta}\cdot 2\sec\theta\tan\theta\,d\theta =2\int \tan^{2}\theta\,d\theta.$$又 $\tan^{2}\theta=\sec^{2}\theta-1$，故
$$2\int \tan^{2}\theta\,d\theta =2\int (\sec^{2}\theta-1)d\theta =2(\tan\theta-\theta)+C.$$由 $x=2\sec\theta$ 得 $\sec\theta=\dfrac{x}{2}$，$\tan\theta=\dfrac{\sqrt{x^{2}-4}}{2}$，且 $\theta=\arccos\dfrac{2}{x}$，故
$$\int \frac{\sqrt{x^{2}-4}}{x}\,dx =\sqrt{x^{2}-4}-2\arccos\frac{2}{x}+C.$$(4) $\displaystyle \int \frac{1}{\sqrt{(x^{2}+1)^{3}}}\,dx$；
解：
写成 $\displaystyle \int (x^{2}+1)^{-\tfrac{3}{2}}dx$。 注意
$$\frac{d}{dx}\left(\frac{x}{\sqrt{x^{2}+1}}\right) =(x^{2}+1)^{-\tfrac{3}{2}}.$$故
$$\int \frac{1}{\sqrt{(x^{2}+1)^{3}}}\,dx =\frac{x}{\sqrt{x^{2}+1}}+C.$$二、提高题
3 求下列不定积分：
(1) $\displaystyle \int \frac{dx}{x\ln x\cdot \ln(\ln x)}$；
解：
先令 $u=\ln x$，则 $du=\dfrac{1}{x}dx$，积分化为
$$\int \frac{1}{u\ln u}\,du.$$再令 $v=\ln u$，则 $dv=\dfrac{1}{u}du$，
$$\int \frac{1}{u\ln u}\,du =\int \frac{1}{v}dv =\ln|v|+C =\ln|\ln u|+C =\ln|\ln(\ln x)|+C.$$(2) $\displaystyle \int \frac{x\ln(1+x^{2})\,dx}{1+x^{2}}$；
解：
令 $u=1+x^{2}$，则 $du=2x\,dx$，$x\,dx=\dfrac{1}{2}du$，
$$\int \frac{x\ln(1+x^{2})}{1+x^{2}}dx =\frac{1}{2}\int \frac{\ln u}{u}du.$$令 $v=\ln u$，$dv=\dfrac{1}{u}du$，
$$\frac{1}{2}\int \frac{\ln u}{u}du =\frac{1}{2}\int v\,dv =\frac{1}{4}v^{2}+C =\frac{1}{4}(\ln(1+x^{2}))^{2}+C.$$(3) $\displaystyle \int \cos^{3}x\,dx$；
解：
$$\int \cos^{3}x\,dx=\int \cos x(1-\sin^{2}x)\,dx.$$令 $u=\sin x$，$du=\cos x\,dx$，
$$\int \cos^{3}x\,dx =\int (1-u^{2})du =u-\frac{u^{3}}{3}+C =\sin x-\frac{\sin^{3}x}{3}+C.$$(4) $\displaystyle \int \frac{1}{3+\sin^{2}x}\,dx$；
解：
令 $t=\tan x$，则 $dx=\dfrac{dt}{1+t^{2}}$，$\sin^{2}x=\dfrac{t^{2}}{1+t^{2}}$，
$$\int \frac{1}{3+\sin^{2}x}dx =\int \frac{1}{3+\dfrac{t^{2}}{1+t^{2}}}\cdot \frac{dt}{1+t^{2}} =\int \frac{dt}{3+4t^{2}}.$$于是
$$\int \frac{dt}{3+4t^{2}} =\frac{1}{4}\int \frac{dt}{t^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} =\frac{1}{4}\cdot \frac{2}{\sqrt{3}}\arctan\frac{2t}{\sqrt{3}}+C =\frac{1}{2\sqrt{3}}\arctan\frac{2\tan x}{\sqrt{3}}+C.$$(5) $\displaystyle \int \sin^{2}x\cos^{4}x\,dx$；
解：
利用降幂公式
$$\sin^{2}x=\frac{1-\cos2x}{2},\quad \cos^{2}x=\frac{1+\cos2x}{2},$$则
$$\sin^{2}x\cos^{4}x =\sin^{2}x(\cos^{2}x)^{2} =\frac{1-\cos2x}{2}\left(\frac{1+\cos2x}{2}\right)^{2}.$$展开并化简可得
$$\sin^{2}x\cos^{4}x =\frac{1}{16}-\frac{1}{16}\cos4x+\frac{1}{32}\cos2x-\frac{1}{32}\cos6x.$$故
$$\int \sin^{2}x\cos^{4}x\,dx =\int\left(\frac{1}{16}-\frac{1}{16}\cos4x+\frac{1}{32}\cos2x-\frac{1}{32}\cos6x\right)dx$$$$=\frac{x}{16}-\frac{1}{16}\cdot\frac{\sin4x}{4} +\frac{1}{32}\cdot\frac{\sin2x}{2} -\frac{1}{32}\cdot\frac{\sin6x}{6}+C$$$$=\frac{x}{16}-\frac{\sin4x}{64}+\frac{\sin2x}{64}-\frac{\sin6x}{192}+C.$$(6) $\displaystyle \int \tan^{5}x\sec^{3}x\,dx$；
解： 写成
$$\int \tan^{5}x\sec^{3}x\,dx =\int \tan^{4}x\tan x\sec^{3}x\,dx =\int (\sec^{2}x-1)^{2}\sec^{2}x(\sec x\tan x\,dx).$$令 $u=\sec x$，则 $du=\sec x\tan x\,dx$，且 $\sec^{2}x=u^{2}$，
$$\int \tan^{5}x\sec^{3}x\,dx =\int (u^{2}-1)^{2}u^{2}du =\int (u^{6}-2u^{4}+u^{2})du$$$$=\frac{u^{7}}{7}-\frac{2u^{5}}{5}+\frac{u^{3}}{3}+C =\frac{\sec^{7}x}{7}-\frac{2}{5}\sec^{5}x+\frac{1}{3}\sec^{3}x+C.$$(7) $\displaystyle \int \cos3x\sin7x\,dx$；
解： 用积化和差公式 $\sin A\cos B=\dfrac{1}{2}[\sin(A+B)+\sin(A-B)]$，
$$\cos3x\sin7x=\frac{1}{2}[\sin(10x)+\sin(4x)].$$故
$$\int \cos3x\sin7x\,dx =\frac{1}{2}\int (\sin10x+\sin4x)dx =-\frac{1}{2}\left(\frac{\cos10x}{10}+\frac{\cos4x}{4}\right)+C =-\frac{\cos10x}{20}-\frac{\cos4x}{8}+C.$$(8) $\displaystyle \int \frac{\ln(\tan x)}{\sin x\cos x}\,dx$；
解： 令 $u=\tan x$，则 $du=\sec^{2}x\,dx=(1+\tan^{2}x)dx$。 又
$$\sin x\cos x=\frac{\tan x}{1+\tan^{2}x}=\frac{u}{1+u^{2}},$$故
$$\frac{dx}{\sin x\cos x} =\frac{1+u^{2}}{u}dx =\frac{1+u^{2}}{u}\cdot\frac{du}{1+u^{2}} =\frac{du}{u}.$$于是
$$\int \frac{\ln(\tan x)}{\sin x\cos x}\,dx =\int \frac{\ln u}{u}du =\frac{1}{2}(\ln u)^{2}+C =\frac{1}{2}[\ln(\tan x)]^{2}+C.$$4 求下列不定积分：
(1) $\displaystyle \int \frac{1-x}{\sqrt{9-4x^{2}}}\,dx$；
解：
分拆为
$$\int \frac{1}{\sqrt{9-4x^{2}}}dx-\int \frac{x}{\sqrt{9-4x^{2}}}dx.$$第一项：令 $t=\dfrac{2x}{3}$，则 $x=\dfrac{3}{2}t$，$dx=\dfrac{3}{2}dt$，且 $\sqrt{9-4x^{2}}=3\sqrt{1-t^{2}}$，
$$\int \frac{1}{\sqrt{9-4x^{2}}}dx =\frac{1}{2}\int \frac{1}{\sqrt{1-t^{2}}}dt =\frac{1}{2}\arcsin t +C =\frac{1}{2}\arcsin\frac{2x}{3}.$$第二项：令 $u=9-4x^{2}$，则 $du=-8x\,dx$，$x\,dx=-\dfrac{1}{8}du$，
$$\int \frac{x}{\sqrt{9-4x^{2}}}dx =-\frac{1}{8}\int u^{-\tfrac{1}{2}}du =-\frac{1}{8}\cdot 2u^{\tfrac{1}{2}}+C =-\frac{1}{4}\sqrt{9-4x^{2}}+C.$$所以
$$\int \frac{1-x}{\sqrt{9-4x^{2}}}\,dx =\frac{1}{2}\arcsin\frac{2x}{3}+\frac{1}{4}\sqrt{9-4x^{2}}+C.$$(2) $\displaystyle \int \frac{x^{3}+1}{(x^{2}+1)^{2}}\,dx$；
解： 将分子改写为
$$x^{3}+1=x(x^{2}+1)-(x-1),$$于是
$$\int \frac{x^{3}+1}{(x^{2}+1)^{2}}dx =\int \frac{x}{x^{2}+1}dx-\int \frac{x-1}{(x^{2}+1)^{2}}dx.$$记
$$A=\int \frac{x}{x^{2}+1}dx=\frac{1}{2}\ln(x^{2}+1),$$$$B=\int \frac{x}{(x^{2}+1)^{2}}dx,\quad C=\int \frac{1}{(x^{2}+1)^{2}}dx.$$对 $B$ 令 $u=x^{2}+1$，$du=2x\,dx$，则
$$B=\frac{1}{2}\int u^{-2}du =-\frac{1}{2u}+C=-\frac{1}{2(x^{2}+1)}+C.$$对 $C$ 利用已知积分结果
$$\int \frac{1}{(x^{2}+1)^{2}}dx =\frac{x}{2(x^{2}+1)}+\frac{1}{2}\arctan x+C.$$于是
$$\int \frac{x^{3}+1}{(x^{2}+1)^{2}}dx =A-(B-C) =A-B+C$$$$=\frac{1}{2}\ln(x^{2}+1)+\frac{1}{2(x^{2}+1)} +\frac{x}{2(x^{2}+1)}+\frac{1}{2}\arctan x+C$$$$=\frac{1}{2}\ln(x^{2}+1)+\frac{x+1}{2(x^{2}+1)}+\frac{1}{2}\arctan x+C.$$(3) $\displaystyle \int \frac{dx}{x+\sqrt{1-x^{2}}}$；
解： 令 $x=\sin t$，$dx=\cos t\,dt$，$\sqrt{1-x^{2}}=\cos t$，分母为 $\sin t+\cos t$，
$$\int \frac{dx}{x+\sqrt{1-x^{2}}} =\int \frac{\cos t}{\sin t+\cos t}dt.$$分子分母同乘共轭 $\cos t-\sin t$，得
$$\frac{\cos t}{\sin t+\cos t} =\frac{\cos t(\cos t-\sin t)}{\cos^{2}t-\sin^{2}t} =\frac{\cos^{2}t-\sin t\cos t}{\cos2t}.$$利用
$$\cos^{2}t=\frac{1+\cos2t}{2},\quad \sin t\cos t=\frac{\sin2t}{2},$$得
$$\frac{\cos^{2}t-\sin t\cos t}{\cos2t} =\frac{1}{2}\left(\sec2t+1-\tan2t\right).$$于是
$$\int \frac{\cos t}{\sin t+\cos t}dt =\frac{1}{2}\int\left(\sec2t+1-\tan2t\right)dt.$$分别积分：
$$\int \sec2t\,dt=\frac{1}{2}\ln|\sec2t+\tan2t|+C,\quad \int \tan2t\,dt=-\frac{1}{2}\ln|\cos2t|+C.$$整理可得
$$\int \frac{\cos t}{\sin t+\cos t}dt =\frac{1}{2}\left(\ln(\sin t+\cos t)+t\right)+C.$$回代 $x=\sin t$，$\cos t=\sqrt{1-x^{2}}$，$t=\arcsin x$，并注意 $\sin t+\cos t=x+\sqrt{1-x^{2}}$，故
$$\int \frac{dx}{x+\sqrt{1-x^{2}}} =\frac{1}{2}\ln\bigl(x+\sqrt{1-x^{2}}\bigr) +\frac{1}{2}\arcsin x+C.$$(4) $\displaystyle \int \frac{1}{\sqrt{1+x-x^{2}}}\,dx$；
解： 配方：
$$1+x-x^{2}=-\left(x^{2}-x-1\right) =\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}.$$令
$$x-\frac{1}{2}=\frac{\sqrt{5}}{2}u,\quad dx=\frac{\sqrt{5}}{2}du,$$则
$$\sqrt{1+x-x^{2}} =\sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}} =\frac{\sqrt{5}}{2}\sqrt{1-u^{2}}.$$因此
$$\int \frac{1}{\sqrt{1+x-x^{2}}}dx =\int \frac{\frac{\sqrt{5}}{2}du}{\frac{\sqrt{5}}{2}\sqrt{1-u^{2}}} =\int \frac{1}{\sqrt{1-u^{2}}}du =\arcsin u +C.$$回代 $u=\dfrac{2x-1}{\sqrt{5}}$，得
$$\int \frac{1}{\sqrt{1+x-x^{2}}}dx =\arcsin\frac{2x-1}{\sqrt{5}}+C.$$(5) $\displaystyle \int \frac{dx}{1+\sqrt{1-x^{2}}}$；
解： 分母有根式，乘以共轭：
$$\frac{1}{1+\sqrt{1-x^{2}}} =\frac{1-\sqrt{1-x^{2}}}{1-(1-x^{2})} =\frac{1-\sqrt{1-x^{2}}}{x^{2}}.$$故
$$\int \frac{dx}{1+\sqrt{1-x^{2}}} =\int \left(\frac{1}{x^{2}}-\frac{\sqrt{1-x^{2}}}{x^{2}}\right)dx =\int x^{-2}dx-\int \frac{\sqrt{1-x^{2}}}{x^{2}}dx.$$第一项
$$\int x^{-2}dx=-\frac{1}{x}.$$对第二项令 $x=\sin t$，则 $dx=\cos t\,dt$，$\sqrt{1-x^{2}}=\cos t$，$x^{2}=\sin^{2}t$，
$$\int \frac{\sqrt{1-x^{2}}}{x^{2}}dx =\int \frac{\cos t}{\sin^{2}t}\cos t\,dt =\int \cot^{2}t\,dt.$$又 $\cot^{2}t=\csc^{2}t-1$，故
$$\int \cot^{2}t\,dt =-\cot t -t +C.$$于是
$$\int \frac{\sqrt{1-x^{2}}}{x^{2}}dx =-\frac{\cos t}{\sin t}-t +C =-\frac{\sqrt{1-x^{2}}}{x}-\arcsin x+C.$$于是原积分
$$\int \frac{dx}{1+\sqrt{1-x^{2}}} =-\frac{1}{x}-\left(-\frac{\sqrt{1-x^{2}}}{x}-\arcsin x\right)+C =\frac{\sqrt{1-x^{2}}-1}{x}+\arcsin x+C.$$(6) $\displaystyle \int \frac{\sqrt{x^{2}-1}}{2x^{2}}\,dx$；
解： 写为
$$\frac{1}{2}\int \frac{\sqrt{x^{2}-1}}{x^{2}}dx.$$对 $x>1$，令 $x=\sec t$，则 $\sqrt{x^{2}-1}=\tan t$，$dx=\sec t\tan t\,dt$，$x^{2}=\sec^{2}t$，
$$\frac{\sqrt{x^{2}-1}}{x^{2}}dx =\frac{\tan t}{\sec^{2}t}\sec t\tan t\,dt =\tan^{2}t\cos t\,dt =\frac{\sin^{2}t}{\cos t}dt.$$故
$$\int \frac{\sqrt{x^{2}-1}}{2x^{2}}dx =\frac{1}{2}\int \frac{\sin^{2}t}{\cos t}dt =\frac{1}{2}\int (\sec t -\cos t)dt.$$于是
$$\int \sec t\,dt=\ln|\sec t+\tan t|+C,\quad \int \cos t\,dt=\sin t+C,$$故
$$\int \frac{\sqrt{x^{2}-1}}{2x^{2}}dx =\frac{1}{2}\left(\ln|\sec t+\tan t|-\sin t\right)+C.$$回代：$\sec t=x$，$\tan t=\sqrt{x^{2}-1}$，$\sin t=\dfrac{\tan t}{\sec t}=\dfrac{\sqrt{x^{2}-1}}{x}$，得
$$\int \frac{\sqrt{x^{2}-1}}{2x^{2}}dx =\frac{1}{2}\ln\bigl|x+\sqrt{x^{2}-1}\bigr| -\frac{1}{2}\cdot \frac{\sqrt{x^{2}-1}}{x}+C.$$三、考研真题
5 $(2004102)$ 已知 $f'(e^{x})=xe^{-x}$，且 $f(1)=0$，求 $f(x)$。
解： 设 $y=e^{x}$，则 $x=\ln y$，$y>0$。题给条件可写为
$$f'(y)=\frac{\ln y}{y},\quad (y>0).$$于是
$$f(x)=\int f'(x)\,dx=\int \frac{\ln x}{x}dx =\frac{1}{2}(\ln x)^{2}+C.$$由 $f(1)=0$ 得 $0=f(1)=\dfrac{1}{2}(\ln1)^{2}+C=C$，故 $C=0$， 于是
$$f(x)=\frac{1}{2}(\ln x)^{2}.$$6 $(2001311)$ 求 $\displaystyle \int \frac{1}{(2x^{2}+1)\sqrt{x^{2}+1}}\,dx$。
解： 令 $x=\sinh u$，则 $dx=\cosh u\,du$，$\sqrt{x^{2}+1}=\sqrt{\sinh^{2}u+1}=\cosh u$，
$$2x^{2}+1=2\sinh^{2}u+1.$$因此
$$\int \frac{1}{(2x^{2}+1)\sqrt{x^{2}+1}}dx =\int \frac{\cosh u\,du}{(2\sinh^{2}u+1)\cosh u} =\int \frac{1}{2\sinh^{2}u+1}du.$$利用恒等式 $\cosh 2u=2\sinh^{2}u+1$，可得
$$\int \frac{1}{2\sinh^{2}u+1}du =\int \operatorname{sech}(2u)\,du.$$而
$$\int \operatorname{sech}z\,dz =2\arctan(\tanh \tfrac{z}{2})+C,$$所以
$$\int \operatorname{sech}(2u)\,du =\frac{1}{2}\cdot 2\arctan(\tanh u)+C =\arctan(\tanh u)+C.$$回代：$x=\sinh u$，$\tanh u=\dfrac{\sinh u}{\cosh u}=\dfrac{x}{\sqrt{x^{2}+1}}$，得
$$\int \frac{1}{(2x^{2}+1)\sqrt{x^{2}+1}}dx =\arctan\frac{x}{\sqrt{x^{2}+1}}+C.$$