431. Find the integrals:
1) $\int \frac{d x}{\sqrt[3]{5 x}}$
2) $\int \frac{d t}{\sqrt{3-4 t^{2}}}$
3) $\int \cos 3 \varphi d \varphi$
4) $\int e^{-\frac{x}{2}} d x$
5) $\int \sin (a x+b) d x$
6) $\int \frac{1}{5 x+4} d x$
Solution.

1) $\int \frac{d x}{\sqrt[3]{5 x}}=\frac{1}{\sqrt[3]{5}} \int x^{-\frac{1}{3}} d x=\frac{1}{\sqrt[3]{5}} \cdot \frac{3}{2} x^{\frac{2}{3}}+C=$ $=\frac{3}{2 \sqrt[3]{5} 5} \sqrt[3]{x^{2}}+C$, according to property III and formula 1, when $u=x, a=-\frac{1}{3}$.
2) $\int \frac{d t}{\sqrt{3-4 t^{2}}}=\frac{1}{2} \int \frac{d(2 t)}{\sqrt{3-(2 t)^{2}}}=\frac{1}{2} \arcsin \frac{2 t}{\sqrt{3}}+C$, according to property III and formula 10, when $u=2 t, a=\sqrt{3}$.
3) We multiply and divide the integral by 3 and bring the factor 3 under the integral sign (according to property III), then under the differential sign:

$$
\int \cos 3 \varphi d \varphi=\frac{1}{3} \int \cos 3 \varphi d(3 \varphi)=\frac{1}{3} \sin 3 \varphi+C
$$

by formula 5, when $u=3 \varphi$.

4) We multiply and divide the integral by -2 and bring the divisor -2 under the integral and differential signs:

$$
\int e^{-\frac{x}{2}} d x=-2 \int e^{-\frac{x}{2}} d\left(-\frac{x}{2}\right)=-2 e^{-\frac{x}{2}}+C
$$

by formula 3, when $u=-\frac{x}{2}$.

5) Multiplying and dividing by $a$ and noting that $a d x=d(a x+b)$, we get

$\int \sin (a x+b) d x=\frac{1}{a} \int \sin (a x+b) d(a x+b)=-\frac{1}{a} \cos (a x+b)+C$, by formula 4, when $u=a x+b$.

6) Multiplying and dividing by 5, we get:

$$
\int \frac{1}{5 x+4} d x=\frac{1}{5} \int \frac{5}{5 x+4} d x=\frac{1}{5} \ln |5 x+4|+C
$$

by formula 2, when $u=5 x+4 ; u^{\prime}=5$.

This integral can also be found differently:

$$
\int \frac{1}{5 x+4} d x=\frac{1}{5} \int \frac{1}{x+\frac{4}{5}} d x=\frac{1}{5} \ln \left|x+\frac{4}{5}\right|+C
$$

by formula 2, when $u=x+\frac{4}{5}, u^{\prime}=1$.

Both results obtained are correct, which can be verified by differentiating them.