591. Calculate the integrals:
1) $\int_{0}^{5} \frac{x d x}{\sqrt{1+3 x}}$
2) $\int_{\ln 2}^{\ln 9} \frac{d x}{e^{x}-e^{-x}}$;
3) $\int_{1}^{\sqrt{3}} \frac{\left(x^{3}+1\right) d x}{x^{2} \sqrt{4-x^{2}}}$;
4) $\int_{0}^{\frac{\pi}{2}} \frac{d x}{2+\cos x}$.
Solution. 1) Introduce a new variable of integration, setting $\sqrt{1+3 x}=t$. From this, we find $x=\frac{t^{2}-1}{3}, d x=\frac{2}{3} t d t$ and the new limits of integration: $t_{1}=1$ when $x_{1}=0, t_{2}=4$ when $x_{2}=5$. Substituting, we get

$$
\begin{aligned}
\int_{0}^{5} \frac{x d x}{\sqrt{1+3 x}} & =\frac{2}{9} \int_{1}^{4}\left(t^{2}-1\right) d t=\left.\frac{2}{9}\left(\frac{t^{3}}{3}-t\right)\right|_{1} ^{4}= \\
& =\frac{2}{9}\left(\frac{64-1}{3}-4+1\right)=4
\end{aligned}
$$

2) Setting $e^{x}=t$, we have $x=\ln t, d x=\frac{d t}{t} ; t_{1}=2$ when $x_{1}=\ln 2$; $t_{2}=3$ when $x_{2}=\ln 3$ and

$$
\begin{aligned}
& \int_{\ln 2}^{\ln 3} \frac{d x}{e^{x}-e^{-x}}=\int_{2}^{3} \frac{d t}{t\left(t-t^{-1}\right)}=\int_{2}^{3} \frac{d t}{t^{2}-1}= \\
&=\left.\frac{1}{2} \ln \frac{t-1}{t+1}\right|_{2} ^{3}=\frac{1}{2}\left(\ln \frac{2}{4}-\ln \frac{1}{3}\right)=\frac{\ln 1.5}{2}
\end{aligned}
$$

3) Setting $x=2 \sin t$, we get: $d x=2 \cos t d t ; t_{1}=\frac{\pi}{6}$ when $x_{1}=1 ; t_{2}=\frac{\pi}{3}$ when $x_{2}=\sqrt{3}$;

$$
\begin{gathered}
\int_{1}^{\sqrt{3}} \frac{\left(x^{3}+1\right) d x}{x^{2} \sqrt{4-x^{2}}}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{8 \sin ^{3} t+1}{4 \sin ^{2} t} d t=2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin t d t+\frac{1}{4} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d t}{\sin ^{2} t}= \\
=-2 \cos t-\left.\frac{1}{4} \operatorname{ctg} t\right|_{\frac{\pi}{6}} ^{\frac{\pi}{3}}=-2\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)-\frac{1}{4}\left(\frac{\sqrt{3}}{3}-\sqrt{3}\right)= \\
=\frac{7}{2 \sqrt{3}}-1 .
\end{gathered}
$$

4) Replacing the variable using the substitution $\operatorname{tg} \frac{x}{2}=z$, we find $\cos x=\frac{1-z^{2}}{1+z^{2}} ; d x=\frac{2 d z}{1+z^{2}}$ (see Chapter IV, § 9); $z_{1}=0$ when $x_{1}=0 ; z_{2}=1$ when $x_{2}=\frac{\pi}{2}$ and

$$
\int_{0}^{\frac{\pi}{2}} \frac{d x}{2+\cos x}=2 \int_{0}^{1} \frac{d z}{z^{2}+3}=\left.\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{z}{\sqrt{3}}\right|_{0} ^{1}=\frac{2}{\sqrt{3}} \cdot \frac{\pi}{6}=\frac{\pi}{3 \sqrt{3}} \cdot
$$