Question

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

$$\begin{gathered} \left(\mathrm{a}\right)x+\frac{x^2}{2}+\frac{x^3}{3}-\log \left|1-x\right|+C \\ \left(\mathrm{b}\right)x+\frac{x^2}{2}-\frac{x^3}{3}-\log \left|1-x\right|+C \\ \left(\mathrm{c}\right)x-\frac{x^2}{2}-\frac{x^3}{3}-\log \left|1+x\right|+C \\ \left(\mathrm{d}\right)x-\frac{x^2}{2}+\frac{x^3}{3}-\log \left|1+x\right|+C \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^3}{x+1}\mathrm{d}x \\ =\int \frac{x^3+1-1}{x+1}\mathrm{d}x \\ =\int \left(\frac{x^3+1}{x+1}-\frac{1}{x+1}\right)\mathrm{d}x \\ =\int \left(\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-\frac{1}{x+1}\right)\mathrm{d}x \\ =\int \left(x^2-x+1-\frac{1}{x+1}\right)\mathrm{d}x \\ =\left(\frac{x^3}{3}-\frac{x^2}{2}+x-\log \left|x+1\right|\right)+C \\ =\frac{x^3}{3}-\frac{x^2}{2}+x-\log \left|x+1\right|+C \\ \mathrm{Therefore},\int \frac{x^3}{x+1}\mathrm{d}x=\frac{x^3}{3}-\frac{x^2}{2}+x-\log \left|x+1\right|+C \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{d}\right). \end{gathered}$$