Question

Integrate with respect to $x$:

$\frac{1}{(x^2+1)(x+3)}$

Answer 1

$\int \frac{1}{(x^2+1)(x+1)}dxlet,\frac{1}{(x^2+1)(x+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x+1}\Rightarrow 1=(A+C)x^2+(3A+B)x+(3B+C)$

comparing both side we get,

$A+C=0⟶\left(1\right)3A+B=0⟶\left(2\right)3B+C=1⟶\left(3\right)$

solving the above equation we get the values,

$A=-\frac{1}{10};B=\frac{3}{10};C=\frac{1}{10}\therefore \int \frac{1}{(x^2+1)(x+1)}dx=\frac{1}{10}\int [\frac{-x+3}{(x^2+1)}+\frac{1}{(x+1)}]dx=\frac{1}{10}\int [\frac{-x}{x^2+1}+\frac{3}{x^2+1}+\frac{1}{(x+1)}]dx=\frac{1}{10}[\frac{1}{2}\log |x^2+1|+3{\tan }^{-1}x+\log |x+1|]+C$