Mathematics

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


ANSWER

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


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
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