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Question Number 74446 by rajesh4661kumar@gmail.com last updated on 24/Nov/19

Answered by Kunal12588 last updated on 24/Nov/19

$$I=\int e^{{tan}^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)dx$$$$lett={tan}^{-1}x$$$$\Rightarrow dt=\frac{dx}{1+x^2}$$$$I=\int e^t(1+tant+{tan}^{2}t)dt$$$$\Rightarrow I=\int e^t(tant+{sec}^{2}t)dt⊛$$$$\Rightarrow I=e^{t}tant+C$$$$\Rightarrow I={xe}^{{tan}^{-1}x}+C$$$$⊛\int e^x[f\left(x\right)+f{}^{\prime }\left(x\right)]dx=e^{x}f\left(x\right)+C$$

Commented by Kunal12588 last updated on 24/Nov/19

$$\int e^x[f\left(x\right)+f{}^{\prime }\left(x\right)]dx$$$$=\int e^{x}f\left(x\right)dx+\int e^{x}f{}^{\prime }\left(x\right)dx$$$$=e^{x}f\left(x\right)-\int e^{x}f{}^{\prime }\left(x\right)dx+\int e^{x}f{}^{\prime }\left(x\right)dx$$$$=e^{x}f\left(x\right)+C$$

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