$∫\frac{x^3}{x+1}dx$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is option (1) → $x-\frac{x^2}{2}+\frac{x^3}{3}-log |1+x|+c$
$I=∫\frac{x^3}{x+1}dx$
so dividing numerator by denominator
$∫\frac{x^3+1-1}{x+1}dx$
$⇒∫\frac{x^3+1}{x+1}dx-∫\frac{1}{x+1}dx$
$x^3+1=(x+1)(x^2-x+1)$
$⇒∫\frac{(x+1)(x^2-x+1)}{x+1}dx-∫\frac{1}{x+1}dx$
$⇒∫x^2dx-∫xdx+∫1dx-∫\frac{1}{x+1}dx$
$⇒\frac{x^3}{3}-\frac{x^2}{2}+x+\log |1+x|+c$