Mathematics

# $\displaystyle \int \dfrac{x^3}{x + 1}$ is equal to

$x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - log |1 + x| + C$

##### SOLUTION
Let $I = \displaystyle \int \dfrac{x^3}{x + 1}dx$
$= \displaystyle \int \dfrac{(x^3+1)-1}{x + 1}dx$
$= \displaystyle \int \left ( (x^2 - x + 1) - \dfrac{1}{(x + 1)} \right ) dx$
$= \dfrac{x^3}{3} - \dfrac{x^2}{2} + x - log | x + 1| + C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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