Question

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

• A. $\frac{2}{a^{2}}\left(x -\frac{b}{a} log\left(ax+b\right)\right)+ C$
• B. $\frac{2}{a^{2}}\left(x -\frac{b}{a} log\left(ax+b\right)\right) -\frac{x^{2}}{a\left(ax+b\right)}+ C$
• C. $\frac{2}{a^{2}}\left(x +\frac{b}{a} log\left(ax+b\right)\right) +\frac{x^{2}}{a\left(ax+b\right)}+ C$
• D. $\frac{2}{a^{2}}\left(x +\frac{b}{a} log\left(ax+b\right)\right) -\frac{x^{2}}{a\left(ax+b\right)}+ C$

Answer 1

Answer: (B)

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

Answer 2

$I= \int\frac{x^{2}}{\left(ax+b\right)^{2}} dx$ Put $ax +b =t \Rightarrow dx =\frac{1}{a} dt$ $\therefore I =\frac{1}{a^{3}}\int\frac{\left(t-b\right)^{2}}{t^{2}} dt =\frac{1}{a^{3}} \int\left(1+\frac{b^{2}}{t^{2}}-\frac{2b}{t}\right) dt$ $= \frac{1}{a^{3}} \left(t -\frac{b^{2}}{t}-2blogt\right) + C$ $\frac{1}{a^{3}}\left(ax + b -\frac{b^{2}}{ax+b}-2b\, log\left(ax+b\right)\right) + C$ $= \frac{2}{a^{2}} \left(x-\frac{b}{a} log\left(ax+b\right)\right)-\frac{x^{2}}{a\left(ax+b\right)} + C$