Question

$\int \frac{x^2}{{(a+bx)}^2}dx,x\ne -\frac{a}{b}$

Answer 1

Given,

$\int \frac{x^2}{{(a+bx)}^2}dx$

integration by parts

$u=x^2,v^{\prime }=\frac{1}{{(a+bx)}^2}$

$=-\frac{x^2}{b(a+bx)}-\int -\frac{2x}{b(a+bx)}dx$

substitute $u=a+bx$, we get,

$=-\frac{x^2}{b(a+bx)}-(-\frac{2}{b^3}(a+bx-a\ln |a+bx|))$

$=-\frac{x^2}{b(a+bx)}+\frac{2}{b^3}(a+bx-a\ln |a+bx|)+C$