Mathematics

Find :
$$\int { \dfrac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 }  } dx$$


SOLUTION
$$\displaystyle\int{\dfrac{3{x}^{2}dx}{{x}^{6}+1}}$$
$$=\displaystyle\int{\dfrac{3{x}^{2}dx}{{\left({x}^{3}\right)}^{2}+1}}$$
$$t={x}^{3}\Rightarrow dt=3{x}^{2}dx$$
$$=\displaystyle\int{\dfrac{dt}{{t}^{2}+1}}$$
$$={\tan}^{-1}{t}+c$$ where $$c$$ is the constant of integration.
$$={\tan}^{-1}{\left({x}^{3}\right)}+c$$ where $$t={x}^{3}$$
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Subjective Hard Published on 17th 09, 2020
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