(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

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