Let I=∫x2(a+bx)2dxPut a+bx = tbdx=dt⇒dx=1bdt G.I=1b∫(t-a)2b2t2dt =1b3∫t2+a2-2att2dt =1b3∫(1+a2t2-2at) dt =1b3t-a2t-2alogt+C Given integral=1b3(a+bx)−2alog(a+bx)−a2a+bx+C

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int \frac{x^{2}}{(a + b x)^{2}} d x =$

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

$\frac{1}{b^{3}} [(a + b x) - 2 a \log (a + b x) - \frac{a^{2}}{a + b x}] + c$

$\frac{1}{b^{3}} [2 a \log (a + b x) + \frac{a^{2}}{a + b x}] + c$

$\frac{1}{b^{3}} [2 a - (a + b x)^{2} + a^{2}] + c$

$\frac{1}{b^{3}} [2 a \log (a + b x) - \frac{a^{2}}{a + b x}] + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$L e t I = \int \frac{x^{2}}{(a + b x)^{2}} d x$

Put a+bx = t

$b d x = d t \Rightarrow d x = \frac{1}{b} d t G . I = \frac{1}{b} \int \frac{{(t - a)}^{2}}{b^{2} t^{2}} d t = \frac{1}{b^{3}} \int \frac{t^{2} + a^{2} - 2 a t}{t^{2}} d t = \frac{1}{b^{3}} \int (1 + \frac{a^{2}}{t^{2}} - \frac{2 a}{t}) d t = \frac{1}{b^{3}} (t - \frac{a^{2}}{t} - 2 a \log t) + C G i v e n i n t e g r a l = \frac{1}{b^{3}} [(a + b x) - 2 a \log |(a + b x)| - \frac{a^{2}}{a + b x}] + C$