430. Find the following integrals and verify the results by differentiation:
1) $\int \frac{d x}{x^{3}}$; 2) $\int \frac{d x}{\sqrt{2-x^{2}}}$; 3) $\int 3^{t} 5^{t} d t$; 4) $\int \sqrt{y+1} d y$; 5) $\int \frac{d x}{2 x^{2}-6}$.
Solution: 1) $\int \frac{d x}{x^{3}}=\int x^{-3} d x=\frac{x^{-2}}{-2}+C=C-\frac{1}{2 x^{2}}$, by formula 1, where $u=x, a=-3$.

Verification. We find the differential of the obtained function and ensure that it equals the integrand:

$$
d\left(C-\frac{1}{2 x^{2}}\right)=-\frac{1}{2}\left(x^{-2}\right)^{\prime} d x=x^{-3} d x=\frac{d x}{x^{3}}
$$

2) $\int \frac{d x}{\sqrt{2-x^{2}}}=\arcsin \frac{x}{\sqrt{2}}+C$, by formula 10, where $u=x$, $a=\sqrt{2}$.

Verification. $d\left(\arcsin \frac{x}{\sqrt{2}}+C\right)=\left(\arcsin \frac{x}{\sqrt{2}}\right)^{\prime} d x=$

$$
=\frac{\left(\frac{x}{\sqrt{2}}\right)^{\prime}}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^{3}}} d x=\frac{d x}{\sqrt{2-x^{2}}}
$$

3) $\int 3^{t} 5^{t} d t=\int 15^{t} d t=\frac{15^{t}}{\ln 15}+C$, by formula 3, where $u=t$. $a=15$.

Verification. $d\left(\frac{15^{t}}{\ln 15}+C\right)=\frac{1}{\ln 15} \cdot 15^{t} \ln 15 d t=15^{t} d t$.
4) $\int \sqrt{y+1} d y:=\int(y+1)^{\frac{1}{2}} d(y+1)=\frac{2}{3}(y+1)^{\frac{3}{2}}+C=$ $=\frac{2}{3} \sqrt{(y+1)^{3}}+C$, by formula 1, where $u=y+1, a=\frac{1}{2}$, since $d(y+1)=d y$.

Verification. $d\left[\frac{2}{3} \sqrt{(y+1)^{3}}+C\right]=\frac{2}{3} \cdot \frac{3}{2}(y+1)^{\frac{1}{2}} d y=\sqrt{y+1} d y$.
5) $\int \frac{d x}{2 x^{2}-6}=\frac{1}{2} \int \frac{d x}{x^{2}-3}=\frac{1}{2} \cdot \frac{1}{2 \sqrt{3}} \ln \left|\frac{x-\sqrt{3}}{x+\sqrt{3}}\right|+C$

according to property III and formula 9, where $u=x, a=\sqrt{3}$.

Verification. $\quad d\left(\frac{1}{4 \sqrt{3}} \ln \left|\frac{x-\sqrt{3}}{x+\sqrt{3}}\right|+C\right)=\frac{1}{4 \sqrt{3}}(\ln |x-\sqrt{3}|-$ $-\ln |x+\sqrt{3}|)^{\prime} d x=\frac{1}{4 \sqrt{3}}\left(\frac{1}{x-\sqrt{3}}-\frac{1}{x+\sqrt{3}}\right) d x=\frac{d x}{2\left(x^{2}-3\right)}$.

Here, to find the derivatives $(\ln |x-\sqrt{3}|)^{\prime}$ and $(\ln |x+$ $+\sqrt{3}|)^{\prime}$, the formula $(\ln |u|)^{\prime}=\frac{u^{\prime}}{u}$ is applied.