$\frac{1}{x^3(1+x^2)}\\ =\frac{A}{x^3}+\frac{B}{x^2}+\frac{C}{x}+\frac{Dx+E}{x^2+1}\\ =\frac{A(x^2+1)+Bx(x^2+1)+Cx^2(x^2+1)+x^3(Dx+E)}{x^3(1+x^2)}\\ =\frac{(C+D)x^4+(B+E)x^3+(A+C)x^2+Bx+A}{x^3(1+x^2)}\\ =>\left\{\begin{array}{l} C+D=0\\ B+E=0 \\ A+C=0\\ B=0\\A=1 \end{array} \right.\\ =>\left\{\begin{array}{l} A=1\\B=0\\C=-1\\ D=1\\E=0 \end{array} \right.\\ =>\frac{1}{x^3(1+x^2)}=\frac{1}{x^3}-\frac{1}{x}+\frac{x}{x^2+1}\\ \int\limits^2_1 {(\frac{1}{x^3}-\frac{1}{x}+\frac{x}{x^2+1})} \, dx\\ =(\frac{-1}{2}\frac{1}{x^2}-ln|x|+\frac{1}{2}ln|x^2+1|)|^2_1\\ =\frac{-1}{8}-ln2+\frac{1}{2}ln5-\frac{-1}{2}+ln1-\frac{1}{2}ln2\\ =\frac{3}{8}-\frac{3}{2}ln2+\frac{1}{2}ln5\\ =\frac{3}{8}-\frac{1}{2}ln2^3+\frac{1}{2}ln5\\ =\frac{3}{8}+\frac{1}{2}(ln5-ln8)\\ =\frac{3}{8}+\frac{1}{2}ln\frac{5}{8}\\$
