Integral of $$$x \left(x + 1\right)^{4}$$$

Find $$$\int x \left(x + 1\right)^{4}\, dx$$$.

Let $$$u=x + 1$$$.

Then $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Thus,

$${\color{red}{\int{x \left(x + 1\right)^{4} d x}}} = {\color{red}{\int{u^{4} \left(u - 1\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{u^{4} \left(u - 1\right) d u}}} = {\color{red}{\int{\left(u^{5} - u^{4}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(u^{5} - u^{4}\right)d u}}} = {\color{red}{\left(- \int{u^{4} d u} + \int{u^{5} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=5$$$:

$$- \int{u^{4} d u} + {\color{red}{\int{u^{5} d u}}}=- \int{u^{4} d u} + {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=- \int{u^{4} d u} + {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=4$$$:

$$\frac{u^{6}}{6} - {\color{red}{\int{u^{4} d u}}}=\frac{u^{6}}{6} - {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\frac{u^{6}}{6} - {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Recall that $$$u=x + 1$$$:

$$- \frac{{\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{6}}{6} = - \frac{{\color{red}{\left(x + 1\right)}}^{5}}{5} + \frac{{\color{red}{\left(x + 1\right)}}^{6}}{6}$$

Therefore,

$$\int{x \left(x + 1\right)^{4} d x} = \frac{\left(x + 1\right)^{6}}{6} - \frac{\left(x + 1\right)^{5}}{5}$$

Simplify:

$$\int{x \left(x + 1\right)^{4} d x} = \frac{\left(x + 1\right)^{5} \left(5 x - 1\right)}{30}$$

Add the constant of integration:

$$\int{x \left(x + 1\right)^{4} d x} = \frac{\left(x + 1\right)^{5} \left(5 x - 1\right)}{30}+C$$

$$$\int x \left(x + 1\right)^{4}\, dx = \frac{\left(x + 1\right)^{5} \left(5 x - 1\right)}{30} + C$$$A