Помогите решить интеграл.

Решите задачу:

$1)\;\; \int \frac{2x\; dx}{x^2-x+7} =[\; x^2-x+7=(x-\frac{1}{2})^2-\frac{1}{4}+7=\\\\=(x-\frac{1}{2})^2+\frac{27}{4}\; ]=\int \frac{2x\; dx}{(x-\frac{1}{2})^2+\frac{27}{4}} =[t=x-\frac{1}{2}\; ,x=t+ \frac{1}{2}\; ,dx=dt\; ]=\\\\=\int \frac{(2t+1)dt}{t^2+ \frac{27}{4} } =\int \frac{2t\, dt}{t^2+\frac{27}{4}}+\int \frac{dt}{t^2+\frac{27}{4}}=[2t\, dt=d(t^2+\frac{27}{4})\; ]=\\\\=ln|t^2+\frac{27}{4}|+\frac{2}{\sqrt{27}}\cdot arctg\frac{2t}{\sqrt{27}}+C=$

$=ln(x^2-x+7)+\frac{2}{\sqrt{27}}arctg\frac{2x-1}{\sqrt{27}}+C$

$2)\; \; \int \frac{x^2+5x}{x^3-2x^2}dx=\int \frac{x(x+5)}{x^2(x-2)} dx=\int \frac{x+5}{x(x-2)}=Q\\\\ \frac{x+5}{x(x-2)} = \frac{A}{x} +\frac{B}{x-2} = \frac{A(x-2)+Bx}{x(x-2)} \\\\x\; \; \; |\; \; A+B=1\\x^0\; \; |\; \; -2A=5\\\\A=-\frac{5}{2}=-2,5\\\\B=1-A=1+\frac{5}{2}=\frac{7}{2}=3,5\\\\Q= -2,5 \int \frac{dx}{x} +3,5\int \frac{dx}{x-2}=-2,5\cdot ln|x|+3,5\cdot ln|x-2|+C$