Assignment 22 Partial integration method for indefinite integral

1.

(1) Twice corner re-division integral

(2)

\[
\int (\arcsin x) ^2 dx = (\arcsin x) ^2 x-2\int \frac{x}{\sqrt{1-x^2}} \arcsin x dx
\]
and
\[
\int \frac{x}{\sqrt{1-x^2}} \arcsin x dx=-\int \arctan x D (\sqrt{1-x^2}) =
-\arcsin x \sqrt{1-x^2}-+\int dx=-\arcsin x \sqrt{1-x^2} +x +c.
\]
So
\[
\int (\arcsin x) ^2 dx= (\arcsin x) ^2 x+2\arcsin x \sqrt{1-x^2} -2x +c.
\]

(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)
\[
\int \frac{\ln x}{x^3} dx =
-\FRAC12 \int \ln x D (\frac{1}{x^2}) =-\frac1{2x^2} \ln x +\frac12 \int \frac{1}{x^3}dx=-\frac1{2x^2} \ln X-\frac14\f Rac{1}{x^2}+c.
\]

(5)
\[
\begin{aligned}
\end{aligned}
\]

(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)
\[
\begin{aligned}
\int x \ln (x+\sqrt{1+x^2}) dx &= \frac12 \int \ln (x+\sqrt{1+x^2}) d (x^2)
\\&= \frac12 \ln (x+\sqrt{1+x^2}) x^2-\frac12 \int x^2 \cdot \frac{1+ \frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}dx
\\&= \frac12 \ln (x+\sqrt{1+x^2}) x^2-\frac12 \int \frac{x^2}{\sqrt{1+x^2}} DX
\\&= \frac12 \ln (x+\sqrt{1+x^2}) x^2-\frac12 \int xd (\sqrt{1+x^2})
\\&=\FRAC12 \ln (x+\sqrt{1+x^2}) x^2-\frac12 x\sqrt{1+x^2} +\frac12 \int \sqrt{1+x^2} DX
\\&=\FRAC12 \ln (x+\sqrt{1+x^2}) x^2-\frac12 x\sqrt{1+x^2} +\frac12 \int \sqrt{1+x^2} DX
\end{aligned}
\]

How to solve the last integral, Classic transform $x =\tan t$, $-\frac \pi 2 \leq t\leq \frac \pi 2$
\[
\int \sqrt{1+x^2} dx =\int \sec^3 t dt.
\]

and
\[
\begin{aligned}
\int \sec^3 T dt = \int \sec t D (\tan t) =\sec t \tan T-\int \tan^2 t \sec t dt=
\\=\sec T \tan T-\int \frac{\sin^2 t}{\cos^3 t} DT
=\sec T \tan T-\int \frac{1-\cos^2 t}{\cos^3 t} DT
\\=\sec T\tan T-\int \sec^3 t dt +\int \sec t DT,
\end{aligned}
\]
The loop appears, so
\[
\int \sec^3 t dt= \frac12 \sec t \tan t + \frac12 \ln |\sec t +\tan t|+c.
\]
(Of course, you can also use the usual method of processing trigonometric functions, the numerator denominator is multiplied by a $\cos T $ to decompose.) )

Assignment 22 Partial integration method for indefinite integral