Решить неопределенный интеграл, заранее большое спасибо

Question 1

Question 2

Решите задачу:

$\int \frac{dx}{cos^2x(3tgx+1)}=\frac{1}{3}\int\frac{d(3tgx+1)}{(3tgx+1)}=\frac{1}{3}ln|3tgx+1|+C\\\\(\frac{1}{3}ln|3tgx+1|+C)'=\frac{1}{3}*\frac{1}{3tgx+1}*\frac{3}{cos^2x}=\frac{1}{cos^2x(3tgx+1)}$

$\int xarcsinxdx=\frac{x^2}{2}arcsinx-\frac{1}{2}\int\frac{x^2dx}{\sqrt{1-x^2}}=\\=\frac{x^2}{2}arcsinx+\frac{x}{4}\sqrt{1-x^2}-\frac{1}{4}arcsinx+C=\\=(\frac{x^2}{2}-\frac{1}{4})arcsinx+\frac{x}{4}\sqrt{1-x^2}+C\\u=arcsinx=\ \textgreater \ du=\frac{dx}{\sqrt{1-x^2}}\\dv=xdx=\ \textgreater \ v=\frac{x^2}{2}$
$\int\frac{x^2dx}{\sqrt{1-x^2}}=-\int\frac{(1-x^2-1)dx}{\sqrt{1-x^2}}=-\int\sqrt{1-x^2}dx+\int\frac{dx}{\sqrt{1-x^2}}=\\-x\sqrt{1-x^2}-\int\frac{x^2dx}{\sqrt{1-x^2}}+arcsinx\\2\int\frac{x^2dx}{\sqrt{1-x^2}}=-x\sqrt{1-x^2}+arcsinx\\\int\frac{x^2dx}{\sqrt{1-x^2}}=-\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}arcsinx$
$\int\sqrt{1-x^2}dx=x\sqrt{1-x^2}+\int\frac{x^2dx}{\sqrt{1-x^2}}\\u=\sqrt{1-x^2}=\ \textgreater \ du=-\frac{xdx}{\sqrt{1-x^2}}\\dv=dx=\ \textgreater \ v=x$
$((\frac{x^2}{2}-\frac{1}{4})arcsinx+\frac{x}{4}\sqrt{1-x^2}+C)'=\\=xarcsinx+\frac{\frac{x^2}{2}-\frac{1}{4}}{\sqrt{1-x^2}}+\frac{1}{4}(\sqrt{1-x^2}-\frac{2x^2}{2\sqrt{1-x^2}})=\\=xarcsinx+\frac{2x^2-1}{4\sqrt{1-x^2}}+\frac{1}{4}*\frac{2-2x^2-2x^2}{2\sqrt{1-x^2}}=xarcsinx+(\frac{2x^2-1+1-2x^2}{4\sqrt{1-x^2}})=\\=xarcsinx$

$\int\frac{x^3dx}{x^4+2}=\frac{1}{4}\int\frac{d(x^4+2)}{x^4+2}=\frac{1}{4}ln|x^4+2|+C\\\\(\frac{1}{4}ln|x^4+2|+C)'=\frac{1}{4}*\frac{4x^3}{x^4+2}=\frac{x^3}{x^4+2}$