Mathematics

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


SOLUTION
Let $$t={\tan}^{-1}{x}\Rightarrow\,dt=\dfrac{1}{1+{x}^{2}}dx$$

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

$$=\displaystyle\int{{e}^{t}dt}$$

$$={e}^{t}+c$$     .........where $$c$$ is the constant of integration

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