Mathematics

The value of $$\int { { e }^{ { \tan ^{ -1 } { x }  } }} \dfrac { \left( 1+x{ +x }^{ 2 } \right)  }{ 1{ +x }^{ 2 } } dx$$ is ?


ANSWER

$${ xe }^{ \tan ^{ -1 }{ x+C } }$$


SOLUTION
$$ \displaystyle \tan^{-1} x = t \quad  \Rightarrow \quad x = \tan t $$

$$ \displaystyle dx. \frac{1}{1+x^2} = dt.$$

$$ \displaystyle I = \int e^t (1+\tan t + \tan^2 t) dt $$

$$ \displaystyle = \int e^t (\tan t+ \sec^2 t ) dt \quad \left[\because \int e^x (f(x) + f'(x). dx = e^x f(x) +c\right] $$ 

$$ \displaystyle = e^t \tan t +c $$  

$$ \displaystyle = e^{\tan^{-1} x} \tan (\tan^{-1}x) + c $$

$$ \displaystyle I = x \, e^{\tan^{-1}x} + c $$ 
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Single Correct Medium Published on 17th 09, 2020
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