Mathematics

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

$\dfrac{x^3}{3}- x+ tan^{-1} x+c$

SOLUTION
$I= \int \dfrac{x^{4}}{1+ x^{2}} dx$
$= \int \dfrac{x^{4}+1-1}{1+x^{2}} dx$
$= \int \dfrac{ (x^{4}-1) }{x^{2}+1}+ \dfrac{1}{1+x^{2}} dx$
$= \int (x^{2}-1)+ \int \dfrac{dx}{x^{2}+1}$
$\dfrac{x^{3}}{3}-x+ \tan^{-1} (x)+ c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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