Integral Calculus Question 280
Question: $ \int_{{}}^{{}}{\frac{x^{2}dx}{{{(a+bx)}^{2}}}}= $
[IIT 1979]
Options:
A) $ \frac{1}{b^{2}}[ x+\frac{2a}{b}\log (a+bx)-\frac{a^{2}}{b}\frac{1}{a+bx} ] $
B) $ \frac{1}{b^{2}}[ x-\frac{2a}{b}\log (a+bx)+\frac{a^{2}}{b}\frac{1}{a+bx} ] $
C) $ \frac{1}{b^{2}}[ x+\frac{2a}{b}\log (a+bx)+\frac{a^{2}}{b}\frac{1}{a+bx} ] $
D) $ \frac{1}{b^{2}}[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{a^{2}}{b}\frac{1}{a+bx} ] $
Show Answer
Answer:
Correct Answer: D
Solution:
Put $ a+bx=t\Rightarrow x=\frac{t-a}{b} $ and $ dx=\frac{dt}{b} $
$ \therefore ,I={{\int_{{}}^{{}}{( \frac{t-a}{b} )}}^{2}}\times \frac{1}{t^{2}}\frac{dt}{b} $
$ =\frac{1}{b^{2}}\int_{{}}^{{}}{( 1-\frac{2a}{t}+a^{2}.{t^{-2}} )},dt=\frac{1}{b^{2}}[ t-2a\log t-\frac{a^{2}}{t} ] $
$ =\frac{1}{b^{2}}[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{a^{2}}{b}\frac{1}{(a+bx)} ] $ .
BETA
