Question

$\int \frac{x^{3}}{x+1} dx $ is equal to

• A. $x+ \frac{x^{2}}{2} +\frac{x^{3}}{3} - log \left|1-x\right| +C$
• B. $x+ \frac{x^{2}}{2} -\frac{x^{3}}{3} - log \left|1-x\right| +C$
• C. $x- \frac{x^{2}}{2} -\frac{x^{3}}{3} - log \left|1+x\right| +C$
• D. $x- \frac{x^{2}}{2} +\frac{x^{3}}{3} - log \left|1+x\right| +C$

Answer 1

Answer: (D)

We have, $I = \int \frac{x^{3}}{x+1} dx = \int\frac{ x^{3} -1 + 1 }{x+1} dx $

$ = \int \left(\frac{\left(x+1\right)\left(x^{2}+1 -x\right)}{x+1} - \frac{1}{x+1}\right) dx $

$ = \int \left( x^{2} +1 -x\right) dx -\int \frac{dx}{x+1} $

$ = \frac{x^{3}}{3} +x - \frac{x^{2}}{2}- log \left|x+1\right| +C$