Find the value of integral ∫x³/(x+1) dx ?

 

Consider the integral

 \[\int \frac{x^{3}}{(x+1)}\; dx = \int \frac{x^{3}+1 -1}{(x+1)}\; dx\]

We have the algebraic identity: $\displaystyle x^{3}+1= (x+1)(x^{2}-x+1)$ .

Hence 

\[\begin{align*} \int \frac{x^{3}+1 -1}{(x+1)}\; dx&=\int \left (\frac{(x^{3}+1)}{x+1} - \frac{1}{(x+1)} \right )\; dx \\ &=\int \frac{x^{3}+1}{x+1}\; dx - \int\frac{1}{x+1} \; dx \\ &=\int \frac{(x+1)(x^{2}-x+1)}{x+1}\; dx -\int \frac{1}{x+1}\; dx\\ &=\int (x^{2}-x+1)\; dx -\int \frac{1}{x+1}\; dx \; \; (\mathbf{Cancelling \; x+1})\\ &= \frac{x^{3}}{3}-\frac{x^{2}}{2}+x- \log \left |x+1 \right |+C\\ \end{align*}\]

And \[\begin{align*} \int \frac{x^{3}+1}{(x+1)}\; dx &= \frac{x^{3}}{3}-\frac{x^{2}}{2}+x- \log \left |x+1 \right |+C\\ \end{align*}\]

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