Question

Integrate $\int \frac{x^3+3x+2}{(x^2+1)(x+1)}.dx$.

Answer 1

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

comparing both side we get,

$A+C=0A+B=3B+C=2$

solving the above equation we get

$A=\frac{1}{2};B=\frac{5}{2};C=-\frac{1}{2}$

Hence,

$\int \frac{x^3+3x+2}{(x^2+1)(x+1)}dx=\int [\frac{\frac{1}{2}x+\frac{5}{2}x}{x^2+1}+\frac{-\frac{1}{2}}{x+1}]dx=\frac{1}{2}\int \frac{x+5}{x^2+1}dx-\frac{1}{2}\int \frac{1}{x+1}dx=\frac{1}{2}[\int \frac{x}{x^2+1}dx+5\int \frac{1}{x^2+1}dx]-\frac{1}{2}\int \frac{1}{x+1}dx=\frac{1}{4}\log |x^2+1|+\frac{5}{2}{\tan }^{-1}x-\frac{1}{2}\log |x+1|+C$