A) \[{{\tan }^{-1}}x+c\]
B) \[{{e}^{{{\tan }^{-1}}x}}+2x+c\]
C) \[{{e}^{{{\tan }^{-1}}x}}+c\]
D) \[{{e}^{{{\tan }^{-1}}x}}-x+c\]
E) \[x{{e}^{{{\tan }^{-1}}x}}+c\]
Correct Answer: E
Solution :
Let \[I=\int{{{e}^{{{\tan }^{-1}}x}}}dx+\int{{{e}^{{{\tan }^{-1}}x}}.\frac{x}{(1+{{x}^{2}})}}dx\] \[=\int{\frac{d}{dx}}(x{{e}^{{{\tan }^{-1}}x}})dx+c\] \[=x{{e}^{{{\tan }^{-1}}x}}+c\]You need to login to perform this action.
You will be redirected in
3 sec