Integral Calculus Question 178
Question: $ \int{\frac{(1+x)}{x{{(1+xe^{x})}^{2}}}dx} $ is
Options:
A) $ \ln | \frac{xe^{x}}{1+xe^{x}} |+\frac{1}{1+xe^{x}}+C $
B) $ (1+xe^{x})+ln| \frac{xe^{x}}{1+xe^{x}} |+C $
C) $ \frac{1}{1+xe^{x}}+ln| xe^{x}(1+xe^{x}) |+C $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ I=\int{\frac{(1+x)}{x{{(1+xe^{x})}^{2}}}dx=\int{\frac{(1+x)e^{x}}{xe^{x}{{(1+xe^{x})}^{2}}}dx}} $ Put $ xe^{x}=t\Rightarrow (e^{x}+xe^{x})dx=dt $ $ I=\int{\frac{dt}{t{{(1+t)}^{2}}}} $ Let $ \frac{1}{t{{(1+t)}^{2}}}=\frac{A}{t}+\frac{B}{1+t}+\frac{D}{{{(1+t)}^{2}}}, $ we get $ A=\frac{1}{{{(1+0)}^{2}}}=1,D=\frac{1}{-1}=-1 $ Equating coefficient of $ t^{2},0=A+B\Rightarrow B=-1 $
$ \therefore I=\int{[ \frac{1}{t}-\frac{1}{1+t}-\frac{1}{{{(1+t)}^{2}}} ]dt} $ $ =ln| t |-ln| 1+t |+\frac{1}{1+t}+C $ $ =ln| \frac{t}{1+t} |+\frac{1}{1+t}+C $ $ =,ln| \frac{xe^{x}}{1+xe^{x}} |+\frac{1}{1+xe^{x}}+C $
BETA
