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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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