1141. Find the integrals:
1) $\int \frac{d x}{3 x^{2}-6 x+5}$
2) $\int \frac{x+3}{x^{2}+4 x-1} d x$
3) $\int \frac{x^{3}-2 x^{2}+x-1}{3+2 x-x^{2}} d x$
4) $\int \frac{2 x-1}{x^{2}-2 x} d x$
Solution. 1) $\int \frac{d x}{3 x^{2}-6 x+5}=\frac{1}{3} \int \frac{d x}{x^{2}-2 x+\frac{5}{3}}=$ $=\frac{1}{3} \int \frac{d x}{(x-1)^{2}+\frac{2}{3}}=\frac{1}{3} \int \frac{d t}{t^{2}+\frac{2}{3}}=\frac{\sqrt{3}}{3 \sqrt{2}} \operatorname{arctg} \frac{t \sqrt{3}}{\sqrt{2}}+C=$ $=\frac{\sqrt{6}}{6} \operatorname{arctg} \frac{\sqrt{6}(x-1)}{2}+C$.

2) Assuming $x+2=t$, we have $x=t-2, d x=d t$, then $\int \frac{x+3}{x^{2}+4 x-1} d x=\int \frac{x+3}{(x+2)^{2}-5} d x=\int \frac{t+1}{t^{2}-5} d t=\int \frac{t}{t^{2}-5} d t+$

$$
\begin{aligned}
& +\int \frac{d t}{t^{2}-5}=\frac{1}{2} \ln \left|t^{2}-5\right|+\frac{1}{2 \sqrt{5}} \ln \left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|+C= \\
& \quad=\frac{1}{2} \ln \left|x^{2}+4 x-1\right|+\frac{1}{2 \sqrt{5}} \ln \left|\frac{x+2-\sqrt{5}}{x+2+\sqrt{5}}\right|+C
\end{aligned}
$$

3) The integrand is an improper fraction. By dividing the numerator by the denominator, we can separate the integer part and represent the fraction as the sum of an integer rational function and a proper fraction:

$$
\begin{gathered}
\frac{x^{3}-2 x^{2}+x-1}{3+2 x-x^{2}}=-x+\frac{4 x}{3+2 x-x^{2}} \\
I=\int\left(-x+\frac{4 x}{3+2 x-x^{2}}\right) d x=-\int x d x+4 \int \frac{x d x}{3+2 x-x^{2}}= \\
=-\frac{x^{2}}{2}+4 I_{1}
\end{gathered}
$$

Let's find the integral

$$
\begin{gathered}
I_{1}=\int \frac{x d x}{4-(x-1)^{2}}=-\int \frac{t+1}{t^{2}-4} d t=-\int \frac{t d t}{t^{2}-4}-\int \frac{d t}{t^{2}-4}= \\
=-\frac{1}{2} \ln \left|t^{2}-4\right|-\frac{1}{4} \ln \left|\frac{t-2}{t+2}\right|+C= \\
=-\frac{1}{2} \ln \left|x^{2}-2 x-3\right|-\frac{1}{4} \ln \left|\frac{x-3}{x+1}\right|+C \\
(x-1=t, x=t+1, d x=d t)[\mathrm{II} ; \text { (1) } \S 3 ;(13) \S 1] .
\end{gathered}
$$

Therefore,

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

4) $I=\int \frac{2 x-1}{x^{2}-2 x} d x=\int \frac{2 x-1}{(x-1)^{2}-1} d x$.

Let $x-1=t$, then $x=t+1, d x=d t$;

$$
\begin{aligned}
I= & \int \frac{2 t+1}{t^{2}-1} d t=\int \frac{2 t}{t^{2}-1} d t+\int \frac{d t}{t^{2}-1}=\ln \left|t^{2}-1\right|+ \\
& +\frac{1}{2} \ln \left|\frac{t-1}{t+1}\right|+C=\ln \left|x^{2}-2 x\right|+ \\
& +\frac{1}{2} \ln \left|\frac{x-2}{x}\right|+C[I I ; \text { (1) } \S 3 ; \text { (13) } 11]
\end{aligned}
$$