A) \[\log |(x-{{e}^{-x}})|+c\]
B) \[\log |(x+{{e}^{-x}})|+c\]
C) \[\log |(1+x{{e}^{x}})|+c\]
D) \[{{(1+x{{e}^{x}})}^{2}}+c\]
E) \[\log |(1-x{{e}^{x}})|+c\]
Correct Answer: C
Solution :
Let \[I=\int{\frac{1+x}{x+{{e}^{-x}}}}dx\] \[=\int{\frac{1+x}{x+\frac{1}{{{e}^{-x}}}}}dx\] \[=\frac{{{e}^{x}}(1+x)}{1+x{{e}^{x}}}dx\] Putting \[1+x{{e}^{x}}=t\] \[\Rightarrow \] \[{{e}^{x}}(1+x)dx=dt\] \[\therefore \] \[I=\int{\frac{1}{t}}dt\] \[=\log t+c\] \[\Rightarrow \] \[I=log(1+x{{e}^{x}}+c)\]
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