1142. Find the integrals:
1) $\int \frac{3 x-5}{\sqrt{x^{2}-4 x+5}} d x$
2) $\int \frac{d x}{\sqrt{1+2 x-3 x^{2}}}$
Solution. 1) $\int \frac{3 x-5}{\sqrt{x^{2}-4 x+5}} d x=\int \frac{3 x-5}{\sqrt{(x-2)^{2}+1}} d x=$ $=\int \frac{3 t+1}{\sqrt{t^{2}+1}} d t=3 \int \frac{t}{\sqrt{t^{2}+1}} d t+\int \frac{d t}{\sqrt{t^{2}+1}}=3 \sqrt{t^{2}+1}+$

$$
\begin{gathered}
+\ln \left(t+\sqrt{t^{2}+1}\right)+C=3 \sqrt{x^{2}-4 x+5}+ \\
\quad+\ln \left(x-2+\sqrt{x^{2}-4 x+5}\right)+C
\end{gathered}
$$

We applied the substitution $t=x-2, d t=d x$.

$$
\begin{aligned}
& \text { 2) } \int \frac{d x}{\sqrt{1+2 x-3 x^{2}}}=\int \frac{d x}{\sqrt{3\left(\frac{1}{3}+\frac{2}{3} x-x^{2}\right)}}= \\
& =\frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{\frac{4}{9}-\left(x-\frac{1}{3}\right)^{2}}}=\frac{1}{\sqrt{3}} \int \frac{d t}{\sqrt{\frac{4}{9}-t^{2}}}= \\
& =\frac{1}{\sqrt{3}} \arcsin \frac{3 t}{2}+C=\frac{1}{\sqrt{3}} \arcsin \frac{3 x-1}{2}+C \\
& \text { where } x-\frac{1}{3}=t, d x=d t \text {. }
\end{aligned}
$$