Mathematics

Evaluate the given integral.
$$\displaystyle\int { \cfrac { x }{ 4+{ x }^{ 4 } }  } dx$$ 


ANSWER

$$\cfrac { 1 }{ 4 } \tan ^{ -1 }{ \cfrac { { x }^{ 2 } }{ 2 } } $$


SOLUTION
Let $$u={x}^{2}\Rightarrow\,du=2x\,dx$$

$$I=\displaystyle\int{\dfrac{x\,dx}{4+{x}^{4}}}$$

$$=\dfrac{1}{2}\displaystyle\int{\dfrac{du}{4+{u}^{2}}}$$

$$=\dfrac{1}{2}\times\dfrac{1}{2}{\tan}^{-1}{\left(\dfrac{u}{2}\right)}+c$$    .......where $$c$$ is the constant of integration.

$$=\dfrac{1}{4}{\tan}^{-1}{\left(\dfrac{{x}^{2}}{2}\right)}+c$$    ......where $$u={x}^{2}$$
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Single Correct Medium Published on 17th 09, 2020
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