Question

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

• A. $e^{{\tan }^{-1}x}+c$
• B. $-e^{{\tan }^{-1}x}+c$
• C. $-e^{{\tan }^{-1}7x}+c$
• D. $\frac{e^{{\tan }^{-1}x}}{x}+c$

Answer 1

The correct option is A $e^{{\tan }^{-1}x}+c$

$I=\int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx$

$d\left(ta{n}^{-1}x\right)=\frac{1}{1+x^2}dx$

$=\int e^{ta{n}^{-1}x}\cdot d\left(ta{n}^{-1}x\right)$

$ta{n}^{-1}x=t$

$=\int e^{t}dt=e^t+c$

$=e^{ta{n}^{-1}x}+c$

$\int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx=e^{ta{n}^{-1}x}+c$