475. Find the integrals:
1) $\int x \cos x d x$
2) $\int \frac{\ln x}{x^{3}} d x$
3) $\int x \operatorname{arctg} x d x$
4) $\int \arcsin x d x$
5) $\int x^{2} e^{3 x} d x$
6) $\int e^{-x} \cos \frac{x}{2} d x$
Solution. 1) Let $u=x, d v=\cos x d x$, then we find: $d u=d x$, $v=\int \cos x d x=\sin x$. Substituting into formula (*), we get $\int x \cos x d x=x \sin x-\int \sin x d x=x \sin x+\cos x+C$.

2) Let $u=\ln x, d v=\frac{d x}{x^{3}}$, then $d u=\frac{d x}{x}, v=\int \frac{d x}{x^{3}}=$ $=\int x^{-3} d x=-\frac{1}{2 x^{2}}$. Substituting into formula (*), we find $\int \frac{\ln x}{x^{3}} d x=-\frac{\ln x}{2 x^{2}}-\int-\frac{1}{2 x^{2}} \cdot \frac{d x}{x}=-\frac{\ln x}{2 x^{2}}-\frac{1}{4 x^{2}}+C=C-\frac{1+2 \ln x}{4 x^{2}}$.
3) Let $u=\operatorname{arctg} x, d v=x d x$, then $d u=\frac{d x}{1+x^{2}}, v=\int x d x=$ $=\frac{x^{2}}{2}$. By formula (*), we get

$$
I=\int x \operatorname{arctg} x d x=\frac{x^{2}}{2} \operatorname{arctg} x-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x
$$

We find the last integral separately:

$$
\int \frac{x^{2}}{1+x^{2}} d x=\int \frac{1+x^{2}-1}{1+x^{2}} d x=\int\left(1-\frac{1}{1+x^{2}}\right) d x=x-\operatorname{arctg} x
$$

Substituting this result into equation (1), we have $I=\frac{x^{2}}{2} \operatorname{arctg} x-\frac{x}{2}+\frac{1}{2} \operatorname{arctg} x+C=C-\frac{x}{2}+\frac{x^{2}+1}{2} \operatorname{arctg} x$.

4) Let $u=\arcsin x, \quad d v=d x$, then we find: $d u=\frac{d x}{\sqrt{1-x^{2}}}$, $v=\int d x=x$. By formula (*) we get

$$
J=\int \arcsin x d x=x \arcsin x-\int \frac{x d x}{\sqrt{1-x^{2}}}
$$

We find the last integral by transforming it to formula 1:

$$
\begin{aligned}
& \int \frac{x d x}{\sqrt{ }} \frac{\overline{1-x^{2}}}{2}-\frac{1}{2} \int\left(1-x^{2}\right)^{-\frac{1}{2}}(-2 x d x)= \\
= & -\frac{1}{2} \int\left(1-x^{2}\right)^{-\frac{1}{2}} d\left(1-x^{2}\right)=-\left(1-x^{2}\right)^{\frac{1}{2}}
\end{aligned}
$$

Substituting into equation (2), we have

$$
J=x \operatorname{arc} \sin x+\sqrt{1-x^{2}}+C
$$

5) Let $u=x^{2}, d v=e^{3 x} d x$, then $d \iota=2 x d x, v=\int e^{3 x} d x=$ $=\frac{1}{3} \int e^{3 x} d(3 x)=\frac{1}{3} e^{3 x}$. By formula (*) we find

$$
I=\int x^{2} e^{3^{x}} d x=\frac{x}{3} e^{3 x}-\frac{2}{3} \int x e^{3 x} d x
$$

We apply the integration by parts formula again to the last integral. Let $u=x, d v=e^{3 x} d x$, then $d u=d x$, $v=\int e^{3 x} d x=\frac{1}{3} e^{3 x}$. By formula (*) we get

$$
\int x e^{3 x} d x=\frac{x}{3} e^{3 x}-\frac{1}{3} \int e^{3 x} d x=\frac{x}{3} e^{3 x}-\frac{1}{9} e^{3 x}
$$

Find the integrals: