488. Find the integrals:
1) $\int \frac{d x}{x^{2}+4 x+8}$
2) $\int \frac{7-8 x}{2 x^{3}-3 x+1} d x$;
3) $\int \frac{3 x-2}{x^{2}+6 x+9} d x$;
4) $\int \frac{6 x^{3}-7 x^{2}+3 x-1}{2 x-3 x^{2}} d x$.
Solution. 1) By completing the square of the quadratic trinomial $x^{2}+4 x+8=(x+2)^{2}+4$, writing $d(x+2)$ instead of $d x$ and integrating, we get

$$
\int \frac{d x}{x^{2}+4 x+8}=\int \frac{d(x+2)}{(x+2)^{2}+4}=\frac{1}{2} \operatorname{arctg} \frac{x+2}{2}+C,
$$

using formula 8, with $u=x+2, a=2$.

2) Completing the square of the quadratic trinomial

$$
\begin{gathered}
2 x^{2}-3 x+1=2\left(x^{2}-\frac{3}{2} x+\frac{1}{2}\right)=2\left[\left(x-\frac{3}{4}\right)^{2}+\frac{1}{2}-\frac{9}{16}\right]= \\
=2\left[\left(x-\frac{3}{4}\right)^{2}-\frac{1}{16}\right] \\
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\end{gathered}
$$

and substituting the variable $x$ with $x-\frac{3}{4}=t$. Then we get: $d x=d t$

$$
I=\int \frac{(7-8 x) d x}{2 x^{2}-3 x+1}=\frac{1}{2} \int \frac{1-8 t}{t^{2}-\frac{1}{16}} d t
$$

Next, we decompose the obtained integral into two integrals, corresponding to the two terms in the numerator, and find them using formulas:

$$
I=\frac{1}{2} \int \frac{d t}{t^{2}-\frac{1}{16}}-2 \int \frac{2 t d t}{t^{2}-\frac{1}{16}}=\frac{1}{2} \cdot \frac{1}{2 \cdot \frac{1}{4}} \ln \left|\frac{t-\frac{1}{4}}{t+\frac{1}{4}}\right|-2 \ln \left|t^{2}-\frac{1}{16}\right|+C . *
$$

Returning to the variable $x$, we finally get

$$
I=\ln \left|\frac{x-1}{x-0.5}\right|-2 \ln \left|x^{2}-1.5 x+0.5\right|+C
$$

3) Completing the square $x^{2}+6 x+9=(x+3)^{2}$, introducing a new variable $t=x+3$; then we get $d x=d t$ and

$$
\begin{gathered}
\int \frac{(3 x-2) d x}{x^{2}+6 x+9}=\int \frac{3 t-11}{t^{2}} d t=\int\left(\frac{3}{t}+\frac{11}{t^{2}}\right) d t=3 \int \frac{d t}{t}-11 \int t^{-2} d t= \\
=3 \ln |t|+11 t^{-1}+C=3 \ln |x+3|+\frac{11}{x+3}+C
\end{gathered}
$$

4) First, we separate the integer part from the improper fraction under the integral, dividing the numerator by the denominator:

$$
\frac{6 x^{3}-7 x^{2}+3 x-1}{2 x-3 x^{2}}=-2 x+1+\frac{x-1}{2 x-3 x^{2}},
$$

then we integrate each term separately:

$$
\begin{gathered}
J=\int \frac{6 x^{8}-7 x^{2}+3 x-1}{2 x-3 x^{2}} d x=-2 \int x d x+\int d x+\int \frac{(x-1) d x}{2 x-3 x^{2}}= \\
=-x^{2}+x+J_{1}
\end{gathered}
$$

The integral $J_{1}$ is transformed using formulas 2 and 9:

$$
\begin{gathered}
J_{1}=-\frac{1}{3} \int \frac{(x-1) d x}{x^{2}-\frac{2}{3} x}=-\frac{1}{3} \int \frac{d x}{x-\frac{2}{3}}+\frac{1}{3} \int \frac{d\left(x-\frac{1}{3}\right)}{\left(x-\frac{1}{3}\right)^{2}-\frac{1}{9}}= \\
=-\frac{1}{3} \ln \left|x-\frac{2}{3}\right|+\frac{1}{3} \cdot \frac{1}{2 \cdot \frac{1}{3}} \ln \left|\frac{x-\frac{1}{3}-\frac{1}{3}}{x-\frac{1}{3}+\frac{1}{3}}\right|= \\
=-\frac{1}{3} \ln \left|x-\frac{2}{3}\right|+\frac{1}{2} \ln \left|\frac{x-\frac{2}{3}}{x}\right|
\end{gathered}
$$

* The first integral is by formula 9 with $u=t, a=\frac{1}{4}$, the second integral is by formula 2 with $u=t^{2}-\frac{1}{16}$.

Finally, we get

$$
J=C-x^{2}+x+\frac{1}{6} \ln \left|x-\frac{2}{3}\right|-\frac{1}{2} \ln |x|
$$