Integral Calculus Question 360
Question: $ \int_{{}}^{{}}{\frac{xe^{x}}{{{(1+x)}^{2}}}dx=} $
[MP PET 1997; UPSEAT 2001; RPET 2002]
Options:
A) $ \frac{{e^{-x}}}{1+x}+c $
B) $ -\frac{{e^{-x}}}{1+x}+c $
C) $ \frac{e^{x}}{1+x}+c $
D) $ -\frac{e^{x}}{1+x}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{xe^{x}}{{{(1+x)}^{2}}},dx=\int_{{}}^{{}}{\frac{(x+1-1)}{{{(1+x)}^{2}}}e^{x}dx}} $ $ =\int_{{}}^{{}}{e^{x}( \frac{1}{1+x}-\frac{1}{{{(1+x)}^{2}}} ),dx}=\frac{e^{x}}{1+x}+c $ .
BETA
