Mathematics

$$\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 }  } dx=$$


ANSWER

$$\dfrac { 1 }{ \sqrt { 2 } } {\tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$$


SOLUTION
Given,

$$\int \dfrac{x^2+1}{x^4+1}dx$$

$$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2}dx$$..............dividing by $${x^2}$$

$$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2-2+2}dx$$

$$=\int \dfrac{\dfrac{1}{x^2}+1}{\left ( x-\dfrac{1}{x} \right )^2+2}dx$$

$$ x-\dfrac{1}{x}=t\rightarrow dt=\dfrac{1}{x^2}+1 dx$$

$$=\int \dfrac{1}{t^2+2}dt$$

$$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{t}{\sqrt 2}+c$$

$$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x-\dfrac{1}{x}}{\sqrt 2}+c$$

$$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x^2-1}{\sqrt 2x}+c$$
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Single Correct Medium Published on 17th 09, 2020
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