Алгебра, вычисление интегралов

Решите задачу:

$1)\ \int6^{5x+2}dx=\frac15\int6^{5x+2}d(5x+2)=\frac{1}{5}*\frac{1}{ln6}*6^{5x+2}+C=\frac{6^{5x+2}}{5ln6}+C \\ \\ 3)\ \int xe^{-x^2}dx=-\frac12 \int e^{-x^2}d(-x^2)=-\frac12 e^{-x^2}+C \\ \\ 4)\ \int {\frac{\sqrt[4]{lnx^2}}{x}}dx=\int {\frac{\sqrt[4]{2lnx}}{x}}dx=\sqrt[4]2\int {\frac{\sqrt[4]{lnx}}{x}}dx=\sqrt[4]2\int {\sqrt[4]{lnx}}d(lnx)=\sqrt[4]2\int (lnx)^{\frac14}d(lnx)= \\ =\sqrt[4]2*\frac{(lnx)^{\frac14+1}}{\frac14+1}+C=\frac{4\sqrt[4]{2}}{5}ln^{\frac54}x+C$

$2)\ \int \frac{x-1}{\sqrt{2x-x^2}}dx=\int \frac{x-1}{\sqrt{x(2-x)}}dx=[u=1-x;\ x=1-u;\ dx=-du]= \\ \\ =-\int \frac{1-u-1}{\sqrt{(1-u)(2-(1-u))}}du=\int \frac{udu}{\sqrt{(1-u)(1+u)}}=\int \frac{udu}{\sqrt{1-u^2}}=-\frac{1}{2}\int \frac{d(1-u^2)}{\sqrt{1-u^2}}= \\ \\ =-\sqrt{1-u^2}+C=-\sqrt{1-(1-x)^2}+C=-\sqrt{1-(1-2x+x^2)}+C= \\ \\ =-\sqrt{2x-x^2}+C$