Mathematics

# Evaluate the given integral.$\displaystyle \int { { e }^{ x } } \left( \tan ^{ -1 }{ x } +\cfrac { 1 }{ 1+{ x }^{ 2 } } \right) dx$

##### SOLUTION
$I=\displaystyle\int{{e}^{x}\left({\tan}^{-1}{x}+\dfrac{1}{1+{x}^{2}}\right)dx}$

$\Rightarrow\,I=\displaystyle\int{{e}^{x}{\tan}^{-1}{x}\,dx}+\displaystyle\int{\dfrac{{e}^{x}}{1+{x}^{2}}dx}$

Let $u={e}^{x}\Rightarrow\,du={e}^{x}\,dx$

$dv=\dfrac{dx}{1+{x}^{2}}\Rightarrow\,v={\tan}^{-1}{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$......by parts formula

$\Rightarrow\,I=\displaystyle\int{{e}^{x}{\tan}^{-1}{x}\,dx}+{e}^{x}{\tan}^{-1}{x}-\displaystyle\int{{e}^{x}{\tan}^{-1}{x}\,dx}+c$

$\therefore\,I={e}^{x}{\tan}^{-1}{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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