1.

(1) 二倍角再分部积分

(2)

\[
\int (\arcsin x)^2 dx = (\arcsin x)^2 x - 2\int \frac{x}{\sqrt{1-x^2}} \arcsin x dx
\]
而
\[
\int \frac{x}{\sqrt{1-x^2}} \arcsin x dx= - \int \arctan x d(\sqrt{1-x^2})=
- \arcsin x \sqrt{1-x^2}-+\int dx=-\arcsin x \sqrt{1-x^2} +x +C.
\]
因此,
\[
\int (\arcsin x)^2 dx=(\arcsin x)^2 x+2\arcsin x \sqrt{1-x^2} -2x +C.
\]

(3)
\[
\int x \tan^2 x dx = \int x(\sec^2 x-1) dx
= \int x d(\tan x) -\frac12 x^2
=
x\tan x +\ln|\cos x| -\frac12 x^2 +C.
\]

(4)
\[
\int \frac{\ln x}{x^3} dx =
-\frac12 \int \ln x d(\frac{1}{x^2})= - \frac1{2x^2} \ln x +\frac12 \int \frac{1}{x^3}dx= - \frac1{2x^2} \ln x - \frac14\frac{1}{x^2}+C.
\]

(5)
\[
\begin{aligned}
\int \sin x \ln \tan x dx = -\int \ln \tan x d(\cos x) =-\ln \tan x \cos x + \int \csc x dx
\\=-\ln( \tan x) \cos x +\ln|\csc x -\cot x| +C.
\end{aligned}
\]

(6)
\[
\begin{aligned}
\int \frac{x\arctan x}{\sqrt{1+x^2}} dx= \int \arctan x d( \sqrt{1+x^2} ) =\arctan x\sqrt{1+x^2} -\int \frac{dx}{\sqrt{1+x^2}}
\\=\arctan x \sqrt{1+x^2} -\ln |x+ \sqrt{1+x^2}|+C.
\end{aligned}
\]

(7)
\[
\begin{aligned}
\int x \ln (x+\sqrt{1+x^2}) dx &= \frac12 \int \ln (x+\sqrt{1+x^2}) d(x^2)
\\&= \frac12 \ln(x+\sqrt{1+x^2}) x^2 -\frac12 \int x^2 \cdot \frac{1+ \frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}dx
\\&= \frac12 \ln(x+\sqrt{1+x^2}) x^2- \frac12 \int \frac{x^2}{\sqrt{1+x^2}} dx
\\&= \frac12 \ln(x+\sqrt{1+x^2}) x^2- \frac12 \int xd (\sqrt{1+x^2})
\\&=\frac12 \ln(x+\sqrt{1+x^2}) x^2- \frac12 x\sqrt{1+x^2} +\frac12 \int \sqrt{1+x^2} dx
\\&=\frac12 \ln(x+\sqrt{1+x^2}) x^2- \frac12 x\sqrt{1+x^2} +\frac12 \int \sqrt{1+x^2} dx
\end{aligned}
\]

最后一个积分如何求解呢，经典变换 $x=\tan t$, $-\frac \pi 2 \leq t\leq \frac \pi 2$ 则
\[
\int \sqrt{1+x^2} dx =\int \sec^3 t dt.
\]

而
\[
\begin{aligned}
\int \sec^3 t dt = \int \sec t d(\tan t) =\sec t \tan t - \int \tan^2 t \sec t dt=
\\=\sec t \tan t - \int \frac{\sin^2 t}{\cos^3 t} dt
=\sec t \tan t - \int \frac{1-\cos^2 t}{\cos^3 t} dt
\\=\sec t\tan t - \int \sec^3 t dt +\int \sec t dt,
\end{aligned}
\]
出现循环, 因此
\[
\int \sec^3 t dt= \frac12 \sec t \tan t + \frac12 \ln |\sec t +\tan t|+C.
\]
(当然也可以用处理三角函数的常用办法，分子分母同乘以一个 $\cos t $ 再去分解。)

作业 22 不定积分的分部积分法