Question

The value of ${\int }_0^1\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}dx$ is

Answer 1

${\int }_0^1\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}dx{\int }_0^1\frac{2x^2+4x+4-x-1}{(x+1)(x^2+2x+2)}dx={\int }_0^1\frac{2(x^2+2x+2)dx}{(x+1)(x^2+2x+2)}-{\int }_0^1\frac{(x+1)dx}{(x+1)(x^2+2x+2)}={\int }_0^1\frac{2}{x+1}dx-{\int }_0^1\frac{dx}{x^2+2x+1+1}$

$={\left[2\ln |x+1|\right]}_0^1-{\int }_0^1\frac{dx}{1+{(x+1)}^2}={[2\ln 2-2\ln 1-{\tan }^{-1}(x+1)]}_0^1=2\ln 2-{\tan }^{-1}2+{\tan }^{-1}1=2\ln 2-{\tan }^{-1}\left[\frac{2-1}{1+2}\right]=2\ln 2-{\tan }^{-1}\frac{1}{3}+C$