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$\it\displaystyle\int\frac{x+arctgx}{1+x^2}dx=\int\frac{x}{1+x^2}dx+\int\frac{arctgx}{1+x^2}dx=\\=\frac{1}{2}\int\frac{d(1+x^2)}{1+x^2}+\int arctgx\ d(arctgx)=\frac{1}{2}ln|1+x^2|+\frac{arctg^2x}{2}+C$

$\it\displaystyle\int\frac{x-7}{x^2+4x+13}dx=\frac{1}{2}\int\frac{2x+4-18}{x^2+4x+13}dx=\\\\\\=\frac{1}{2}\int\frac{2x+4}{x^2+4x+13}dx-9\int\frac{dx}{x^2+4x+13}=\\\\\\=\frac{1}{2}\int\frac{d(x^2+4x+13)}{x^2+4x+13}-9\int\frac{d(x+2)}{(x+2)^2+9}=\\\\\\=\frac{1}{2}ln|x^2+4x+13|-3arctg\frac{x+2}{3}+C\\\\\\(x^2+4x+13)'=2x+4$

$\it\displaystyle\int\frac{3x^2+2x-1}{x(x+1)^2}dx=-\int\frac{dx}{x}+4\int\frac{d(x+1)}{x+1}=\\\\\\=-ln|x|+4ln|x+1|+C=ln|\frac{(x+1)^4}{x}|+C\\\\\frac{3x^2+2x-1}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}=-\frac{1}{x}+\frac{4}{x+1}\\3x^2+2x-1=A(x^2+2x+1)+B(x^2+x)+Cx\\x^2|3=A+B\rightarrow B=4\\x^1|2=2A+B+C\rightarrow C=0\\x^0|-1=A$