Вычислить интеграл под 2 и 4

2)
$\int_{1}^{2}\left ( x+\frac{1}{x}\right )^2{\mathrm{d}x}=\int_{1}^{2}\left ( x^2+2+\frac{1}{x^2}\right ){\mathrm{d}x}=\frac{x^3}{3}|_{1}^{2}+2x|_{1}^{2}-\frac{1}{x}|_{1}^{2}=\frac{1}{3}(8-1)+2(2-1)-(\frac{1}{2}-1)=\frac{7}{3}+2+\frac{1}{2}=\frac{29}{6}\\$ 4)
$\\\int_{-2}^{-1}\frac{4}{x^2}\left( 1-\frac{2}{x}\right ){\mathrm{d}x}=\int_{-2}^{-1}\left(\frac{4}{x^2}-\frac{8}{x^3}\right ){\mathrm{d}x}=-\frac{4}{x}|_{-2}^{-1}+\frac{4}{x^2}|_{-2}^{-1}=-4(-1-\frac{1}{2})+4(1-\frac{1}{4})=2+3=5\\$