Integral Calculus Question 307
Question: $ \int_{{}}^{{}}{\frac{2x}{{{(2x+1)}^{2}}}dx=} $
[DSSE 1985]
Options:
A) $ \frac{1}{2}\log (2x+1)+\frac{1}{2(2x+1)}+c $
B) $ \frac{1}{2}\log (2x+1)-\frac{1}{2(2x+1)}+c $
C) $ 2\log (2x+1)+\frac{1}{2(2x+1)}+c $
D) $ 2\log (2x+1)-\frac{1}{2(2x+1)}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{2x}{{{(2x+1)}^{2}}},dx=\int_{{}}^{{}}{\frac{2x+1-1}{{{(2x+1)}^{2}}},dx}} $ $ =\int_{{}}^{{}}{\frac{1}{(2x+1)},dx}-\int_{{}}^{{}}{{{(2x+1)}^{-2}},dx} $ $ =\frac{1}{2}\log (2x+1)+\frac{1}{2(2x+1)}+c $ .
BETA
