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Question

$$\displaystyle \int e^{\tan ^{-1}x}(1+x+x^{2})  d(\cot ^{-1}x)$$ is equal to


A
etan1x+c
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B
etan1x+c
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C
xetan1x+c
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D
xetan1x+c
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Solution

The correct option is C $$\displaystyle -xe ^{\displaystyle \tan ^{-1}x}+c$$
$$\displaystyle \int e^{\tan ^{-1}x}(1+x+x^{2})  d(\cot ^{-1}x)$$
$$I=\displaystyle \int e^{\tan^{-1}x}(1+x+x^{2})\left ( -\left ( \frac{1}{1+x^{2}} \right )dx \right )$$
$$=-\displaystyle \int e^{\tan^{-1}x}\left ( 1+\frac{x}{1+x^{2}} \right )dx$$
$$=-\displaystyle \int d(xe^{\tan^{-1}x})$$
$$=-\displaystyle xe^{\tan^{-1}x}+C$$

Mathematics

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