A) \[\frac{{{e}^{x}}}{x+1}+c\]
B) \[\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c\]
C) \[\frac{-{{e}^{x}}}{(x+1)}+c\]
D) None of these
Correct Answer: A
Solution :
\[\int{\frac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx\] \[=\int{{{e}^{x}}}\left[ \frac{x+1-1}{{{(x+1)}^{2}}} \right]dx\] \[=\int{{{e}^{x}}}\left[ \frac{1}{x+1}-\frac{1}{{{(x+1)}^{2}}} \right]dx\] \[={{e}^{x}}.\frac{1}{x+1}+c\] \[[\because \int{{{e}^{ax+b}}[af(x)+f'(x)]dx={{e}^{ax+b}}f(x)+c}]\]
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