Integral of $$$\left(u + v\right)^{c - 1}$$$ with respect to $$$u$$$

Find $$$\int \left(u + v\right)^{c - 1}\, du$$$.

Let $$$w=u + v$$$.

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

Therefore,

$${\color{red}{\int{\left(u + v\right)^{c - 1} d u}}} = {\color{red}{\int{w^{c - 1} d w}}}$$

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

$${\color{red}{\int{w^{c - 1} d w}}}={\color{red}{\frac{w^{\left(c - 1\right) + 1}}{\left(c - 1\right) + 1}}}={\color{red}{\frac{w^{c}}{c}}}$$

Recall that $$$w=u + v$$$:

$$\frac{{\color{red}{w}}^{c}}{c} = \frac{{\color{red}{\left(u + v\right)}}^{c}}{c}$$

Therefore,

$$\int{\left(u + v\right)^{c - 1} d u} = \frac{\left(u + v\right)^{c}}{c}$$

Add the constant of integration:

$$\int{\left(u + v\right)^{c - 1} d u} = \frac{\left(u + v\right)^{c}}{c}+C$$

$$$\int \left(u + v\right)^{c - 1}\, du = \frac{\left(u + v\right)^{c}}{c} + C$$$A