A) \[\frac{1}{{{(x+4)}^{2}}}+c\]
B) \[\frac{{{e}^{x}}}{{{(x+4)}^{2}}}+c\]
C) \[\frac{{{e}^{x}}}{x+4}+c\]
D) \[\frac{{{e}^{x}}}{x+3}+c\]
Correct Answer: C
Solution :
\[I=\int{\frac{(x+3){{e}^{x}}}{{{(x+4)}^{2}}}dx}\]\[I=\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}dx}\] \[\Rightarrow I=\int{{{e}^{x}}\,\left( \frac{1}{x+4}-\frac{1}{{{(x+4)}^{2}}} \right)\,dx}\] \[\therefore I=\frac{{{e}^{x}}}{x+4}+c\].
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