A) \[\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c\]
B) \[\frac{{{e}^{x}}}{(x+1)}+c\]
C) \[\frac{-{{e}^{x}}}{(x+1)}+c\]
D) \[\frac{-{{e}^{x}}}{{{(x+1)}^{2}}}+c\]
Correct Answer: A
Solution :
\[\int{{{e}^{x}}\left\{ \frac{x-1}{{{(x+1)}^{3}}} \right\}}dx\] \[=\int{{{e}^{x}}\left\{ \frac{x-1+1-1}{{{(x+1)}^{3}}} \right\}}dx\] \[=\int{{{e}^{x}}\left\{ \frac{1}{{{(x+1)}^{2}}}+\frac{2}{{{(x+1)}^{3}}} \right\}}dx\] \[=\int{\frac{{{e}^{x}}}{{{(x+1)}^{2}}}}dx-2\int{\frac{{{e}^{x}}}{{{(x+1)}^{3}}}}dx\] \[=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+\int{\frac{2{{e}^{x}}}{{{(x+1)}^{3}}}dx}-\int{\frac{2{{e}^{x}}}{{{(x+1)}^{3}}}}dx+c\] \[=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c\]
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