Prove

Question:

$\frac{e^{\tan ^{-1} x}}{1+x^{2}}$

Solution:

Let $\tan ^{-1} x=t$

$\therefore \frac{1}{1+x^{2}} d x=d t$

$\Rightarrow \int \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$

$=e^{\prime}+\mathrm{C}$

$=e^{\tan ^{-1} x}+\mathrm{C}$

 

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