449. Find the integrals:
1) $\int\left(3 x^{2}-2 x+5\right) d x$;
2) $\int \frac{2 x^{2}+x-1}{x^{3}} d x$;
3) $\int\left(1+e^{x}\right)^{2} d x$
4) $\int \frac{2 x+3}{x^{2}-5} d x$;
5) $\int \frac{x^{2}}{x^{2}+1} d x$;
6) $\int \operatorname{tg}^{2} \varphi d \varphi$.
Solution. 1) Integrating each term separately, we get:

$$
\begin{aligned}
& \int\left(3 x^{2}-2 x+5\right) d x=\int 3 x^{2} d x-\int 2 x d x+\int 5 d x=3 \int x^{2} d x- \\
& -2 \int x d x+5 \int d x=3 \cdot \frac{x^{3}}{3}-2 \cdot \frac{x^{2}}{2}+5 x+C=x^{3}-x^{2}+5 x+C
\end{aligned}
$$

by formula 1.

2) We decompose the integrand into terms, dividing the numerator term by term by the denominator. Then we integrate each term separately:

$\int \frac{2 x^{2}+x-1}{x^{3}} d x=2 \int \frac{d x}{x}+\int x^{-2} d x-\int x^{-3} d x=2 \ln |x|-\frac{1}{x}+\frac{1}{2 x^{2}}+C$,

by formulas 2,1.

(3) Squaring and integrating each term, we get

$$
\begin{gathered}
\int\left(1+e^{x}\right)^{2} d x=\int\left(1+2 e^{x}+e^{2 x}\right) d x= \\
=\int d x+2 \int e^{x} d x+\frac{1}{2} \int e^{2 x} d(2 x)=x+2 e^{x}+\frac{1}{2} e^{2 x}+C
\end{gathered}
$$

4) We decompose the integrand fraction into two terms, then integrate using formulas 2 and 9: $\int \frac{2 x+3}{x^{2}-5} d x=\int \frac{2 x d x}{x^{2}-5}+3 \int \frac{d x}{x^{2}-5}=\ln \left|x^{2}-5\right|+\frac{3}{2 \sqrt{5}} \ln \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+C$.
5) Dividing the numerator by the denominator, we exclude the integer part from the improper integrand fraction, then integrate:

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

6) Using the trigonometric formula $1+\operatorname{tg}^{2} \alpha=\sec ^{2} \alpha$, we have:

$$
\int \operatorname{tg}^{2} \varphi d \varphi=\int\left(\sec ^{2} \varphi-1\right) d \varphi=\int \sec ^{2} \varphi d \varphi-\int d \varphi=\operatorname{tg} \varphi-\varphi+C
$$

Find the integrals: