Mathematics

$$ \int { \left( \dfrac { { x }^{ 6 }-1 }{ { x }^{ 2 }+1 }  \right) dx }$$


SOLUTION

Consider the given integral.

$$I=\int{\left( \dfrac{{{x}^{6}}-1}{{{x}^{2}}+1} \right)}dx$$

$$I=\int{\left( {{x}^{4}}-{{x}^{2}}+1-\dfrac{2}{{{x}^{2}}+1} \right)}dx$$

 

We know that

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

 

Therefore,

$$ I=\int{\left( {{x}^{4}}-{{x}^{2}}+1-\dfrac{2}{{{x}^{2}}+1} \right)}dx $$

$$ I=\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{3}}}{3}+x-2{{\tan }^{-1}}x+C $$

 

Hence, this is the answer.

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