Integral Calculus Question 431
Question: $ \int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}e^{x}\ dx=} $
[IIT 1983; MP PET 1990]
Options:
A) $ \frac{-e^{x}}{{{(x+1)}^{2}}}+c $
B) $ \frac{e^{x}}{{{(x+1)}^{2}}}+c $
C) $ \frac{e^{x}}{{{(x+1)}^{3}}}+c $
D) $ \frac{-e^{x}}{{{(x+1)}^{3}}}+c $
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}e^{x}dx=\int_{{}}^{{}}{e^{x}( \frac{(x+1)}{{{(x+1)}^{3}}}-\frac{2}{{{(x+1)}^{3}}} ),dx}} $
$ =\int_{{}}^{{}}{e^{x}( \frac{1}{{{(x+1)}^{2}}}-\frac{2}{{{(x+1)}^{3}}} ),dx}=\frac{e^{x}}{{{(x+1)}^{2}}}+c $ . $ { \therefore ,\frac{d}{dx}( \frac{1}{{{(x+1)}^{2}}} )=-\frac{2}{{{(x+1)}^{3}}} } $
BETA
