Integral Calculus Question 377
Question: $ \int{\frac{(x+3)e^{x}}{{{(x+4)}^{2}}}dx=} $
[Karnataka CET 2000]
Options:
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 $
Show Answer
Answer:
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},( \frac{1}{x+4}-\frac{1}{{{(x+4)}^{2}}} ),dx} $
$ \therefore I=\frac{e^{x}}{x+4}+c $ .
BETA
