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  4. ∫ e tan -1 x(1+x+x2) ⋅ d( cot -1 x) is equal to -

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Q. $\int e^{\tan ^{-1} x}\left(1+x+x^2\right) \cdot d\left(\cot ^{-1} x\right)$ is equal to -

Integrals

Solution:

Correct answer is (c) $-x \cdot e^{\tan ^{-1} x}+c$