458. Find the integrals:
1) $\int \frac{2 x d x}{x^{4}+3}$
2) $\int \frac{\sin x d x}{\sqrt{1+2 \cos x}}$
3) $\int \frac{x d x}{\sqrt[3]{x^{2}+a}}$;
4) $\int \frac{\sqrt{1+\ln x}}{x} d x$
5) $\int \frac{d y}{\sqrt{e^{y}+1}}$
6) $\int \frac{d t}{\sqrt{\left(1-t^{2}\right)^{3}}}$.
Solution. 1) Let $x^{2}=t$; differentiate $2 x d x=d t$, substitute into the integrand, find the new integral, and return to the original variable $x$:

$$
\int \frac{2 x d x}{x^{4}+3}=\int \frac{d t}{t^{2}+3}=\frac{1}{\sqrt{3}} \operatorname{arctg} \frac{t}{\sqrt{3}}+C=\frac{1}{\sqrt{3}} \operatorname{arctg} \frac{x^{2}}{\sqrt{3}}+C
$$

2) Let $1+2 \cos x=t$. Then $-2 \sin x d x=d t$ and

$$
\begin{gathered}
\int \frac{\sin x d x}{\sqrt{1+2 \cos x}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\frac{1}{2} \int t^{-\frac{1}{2}} d t=-\frac{1}{2} \cdot 2 t^{\frac{1}{2}}+C= \\
=C-\sqrt{t}=C-\sqrt{1+2 \cos x}
\end{gathered}
$$

3) Taking the substitution $x^{2}+a=z$, we have $2 x d x=d z$ and

$$
\begin{gathered}
\int \frac{x d x}{\sqrt[3]{x^{2}+a}}=\frac{1}{2} \int \frac{d z}{\sqrt[3]{z}}=\frac{1}{2} \int z^{-\frac{1}{3}} d z=\frac{1}{2} \cdot \frac{3}{2} z^{\frac{2}{3}}+C= \\
=\frac{3}{4} \sqrt[3]{\left(x^{2}+a\right)^{2}}+C
\end{gathered}
$$

4) The substitution $1+\ln x=v$ gives $\frac{d x}{x}=d v$ and

$$
\int \frac{\sqrt{1+\ln x}}{x} d x=\int \sqrt[v]{v} d v=\frac{2}{3} v^{\frac{3}{2}}+C=\frac{2}{3} \sqrt{(1+\ln x)^{3}}+C
$$

5) Taking the substitution $e^{y}+1=t^{2}$, differentiate $e^{y} d y=2 t d t$, determine $d y=\frac{2 t d t}{e^{y}}=\frac{2 t d t}{t^{2}-1}$, substitute into the integrand, integrate, and return to the variable $y$:

$$
\begin{gathered}
\int \frac{d y}{\sqrt{e^{y}+1}}=\int \frac{2 t d t}{t\left(t^{2}-1\right)}=2 \int \frac{d t}{t^{2}-1}=2 \cdot \frac{1}{2} \ln \frac{t-1}{t+1}+C= \\
=\ln \frac{\sqrt{e^{y}+1}-1}{\sqrt{e^{y}+1}+1}+C
\end{gathered}
$$

6) Let $t=\sin \varphi$, then $d t=\cos \varphi d \varphi$,

$$
\int \frac{d t}{\sqrt{\left(1-t^{2}\right)^{3}}}=\int \frac{d \varphi}{\cos ^{2} \varphi}=\operatorname{tg} \varphi+C=\frac{\sin \varphi}{\cos \varphi}+C=\frac{t}{\sqrt{1-t^{2}}}+C
$$

Find the following integrals and verify the results by differentiation: