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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

$\frac{2}{a^{2}} \{\frac{x}{2} - \frac{b}{a} \log (ax + b)\} - \frac{b^{2}}{a^{3} (ax + b)} + C$

$\frac{2}{a^{2}} \{\frac{x}{2} + \frac{b}{a} \log (ax + b)\} + \frac{b^{2}}{a^{3} (ax + b)} + C$

$\frac{2}{a^{2}} \{\frac{x}{2} + \frac{b}{a} \log (ax + b)\} - \frac{b^{2}}{a^{3} (ax + b)} + C$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

put $ax + b = t \Rightarrow dx = \frac{1}{a} dt and x = (\frac{t - b}{a})$

$\begin{aligned} ∴ I = \frac{1}{a^{3}} \int \frac{(t - b)^{2}}{t^{2}} dt = \frac{1}{a^{3}} \int (1 + \frac{b^{2}}{t^{2}} - \frac{2 b}{t}) dt \\ = \frac{1}{a^{3}} (t - \frac{b^{2}}{t} - 2 blog t) + C \end{aligned}$

$\begin{aligned} = \frac{1}{a^{3}} [ax + b - \frac{b^{2}}{ax + b} - 2 blog (ax + b)] + C \\ = \frac{2}{a^{2}} [\frac{x}{2} - \frac{b}{a} \log (ax + b)] - \frac{b^{2}}{a^{3} (ax + b)} + C \end{aligned}$