求不定积分

\[\int\frac{dx}{x(1+\sqrt[3]{x})^2}.\]

Hint.

令 $t=\sqrt[3]{x}$, 则 $x=t^3$, 从而原积分化为

\[
\int\frac{3t^2dt}{t^3(1+t)^2}=\int\frac{3dt}{t(1+t)^2},
\]

这是有理函数的不定积分, 令

\[
\frac{3dt}{t(1+t)^2}=\frac{a}{t}+\frac{bt+c}{(1+t)^2}+\frac{d}{1+t},
\]

当 $d=0$ 时, 可以得 $a=3$, $b=-3$, $c=-6$.

求不定积分

\[\int\ln(\sqrt{1+x}+\sqrt{1-x})dx.\]

求不定积分

\[\int\frac{1}{x(x^6+4)}dx.\]

求不定积分

\[\int\frac{\sqrt{x^2-1}}{x^2}dx.\]

其在 [1,2] 上的定积分见问题1412

证明

\[
\int\sec xdx=\int\frac{1}{\cos x}dx=\ln |\sec x+\tan x|+C.
\]

\[
\int\csc xdx=\int\frac{1}{\sin x}dx=\frac{1}{2}\ln\biggl|\frac{1-\cos x}{1+\cos x}\biggr|+C=\ln\biggl|\tan\frac{x}{2}\biggr|+C.
\]

并求 $\int \sec^2 xdx$, $\int \sec^3 xdx$

[总结]

\[
\begin{aligned}
\int\sec xdx&=\ln|\sec x+\tan x|+C\\
\int\sec^2 xdx&=\tan x+C\\
\int\sec^3 xdx&=\frac{1}{2}\ln|\sec x+\tan x|+\frac{1}{2}\sec x\cdot\tan x+C
\end{aligned}
\]

(1)

\[\int\frac{x^2}{a^2-x^6}dx,\quad (a>0)\]

(2)

\[
\int\frac{\ln\tan x}{\sin x\cos x}dx
\]

(3)

\[
\int\frac{x+\sin x}{1+\cos x}dx
\]

Hint

(1) $x^2dx=\frac{1}{3}dx^3$, 然后令 $t=x^3$.

(2)

\[
\int\frac{\ln\tan x}{\sin x\cos x}dx=\int\frac{\ln\tan x}{\tan x}\cdot\frac{1}{\cos^2 x}dx=\int\frac{\ln\tan x}{\tan x}d\tan x
\]

(3)

\[
\frac{x+\sin x}{1+\cos x}=\frac{(x+\sin x)(1-\cos x)}{\sin^2 x}=\frac{x-x\cos x+\sin x-\sin x\cos x}{\sin^2 x}
\]

因此

\[
\begin{split}
\int\frac{x+\sin x}{1+\cos x}dx&=\int\frac{x}{\sin^2 x}dx-\int\frac{x\cos x}{\sin^2 x}dx+\int\frac{\sin x}{\sin^2 x}dx-\int\frac{\cos x}{\sin x}dx\\
&=-\int x d\cot x-\int\frac{xd\sin x}{\sin^2 x}-\int\frac{d\cos x}{\sin^2 x}-\int\frac{d\sin x}{\sin x}
\end{split}
\]

或者

\[
\begin{split}
\int\frac{x+\sin x}{1+\cos x}dx&=\int\frac{x+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}dx\\
&=\frac{1}{2}\int\frac{x}{\cos^2\frac{x}{2}}dx+\int\tan\frac{x}{2}dx\\
&=\int xd\tan\frac{x}{2}+\int\tan\frac{x}{2}dx\\
&=x\tan\frac{x}{2}-\int\tan\frac{x}{2}dx+\int\tan\frac{x}{2}dx\\
&=x\tan\frac{x}{2}+C
\end{split}
\]

求

\[\int\frac{1}{\sin^2 x\cos x}dx.\]

[Hint]

\[
\int\frac{1}{\sin^2 x\cos x}dx=\int\sec x\csc^2 xdx=-\int\sec x d\cot x
\]

然后用分部积分.