A) \[0\]
B) \[\frac{{{(x+1)}^{2}}}{{{e}^{x}}}+C\]
C) \[\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+C\]
D) \[\frac{{{e}^{x}}}{{{(x-1)}^{2}}}+{{(x+1)}^{2}}+C\]
Correct Answer: C
Solution :
Let \[I=\int{\frac{(x-1){{e}^{x}}}{{{(x+1)}^{3}}}dx}\] \[\Rightarrow \] \[I=\int{\left\{ \frac{x+1-2}{{{(x+1)}^{3}}} \right\}{{e}^{x}}\,\,dx}\] \[=\int{\left\{ \frac{1}{{{(x+1)}^{2}}}-\frac{2}{{{(x+1)}^{3}}} \right\}{{e}^{x}}\,\,dx}\] \[=\int{{{e}^{x}}}\cdot \frac{1}{{{(x+1)}^{2}}}dx-2\int{{{e}^{x}}}\frac{1}{{{(x+1)}^{3}}}dx\] Applying integrating by parts, we get \[=\left( \frac{1}{{{(x+1)}^{2}}}{{e}^{x}}-\int{{{e}^{x}}}\frac{(-2)}{{{(x+1)}^{3}}}dx \right)\] \[-2\int_{{}}^{{}}{{{e}^{x}}}\frac{1}{{{(x+1)}^{3}}}+C\] \[=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+C\] Note We can use the formula \[\int{{{e}^{x}}}[f(x)+f(x)]dx={{e}^{x}}f(x)+C\]
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