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
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