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Q. $\int x^{2}\left(ax + b\right)^{-2} dx$ is equal to

Integrals

Solution:

$I= \int\frac{x^{2}}{\left(ax+b\right)^{2}} dx$

Put $ax +b =t \Rightarrow dx =\frac{1}{a} dt $

$\therefore I =\frac{1}{a^{3}}\int\frac{\left(t-b\right)^{2}}{t^{2}} dt =\frac{1}{a^{3}} \int\left(1+\frac{b^{2}}{t^{2}}-\frac{2b}{t}\right) dt$

$= \frac{1}{a^{3}} \left(t -\frac{b^{2}}{t}-2blogt\right) + C $

$ \frac{1}{a^{3}}\left(ax + b -\frac{b^{2}}{ax+b}-2b\, log\left(ax+b\right)\right) + C $

$= \frac{2}{a^{2}} \left(x-\frac{b}{a} log\left(ax+b\right)\right)-\frac{x^{2}}{a\left(ax+b\right)} + C$