# How do you use substitution to integrate x*(1-x)^n?

Jun 1, 2015

Let $u = 1 - x$ so that $\mathrm{du} = - \mathrm{dx}$ and $x = 1 - u$. Then

$\setminus \int x {\left(1 - x\right)}^{n} \setminus \mathrm{dx} = \setminus \int \left(u - 1\right) {u}^{n} \setminus \mathrm{du} = \setminus \int \left({u}^{n + 1} - {u}^{n}\right) \setminus \mathrm{du}$

$= \setminus \frac{{u}^{n + 2}}{n + 2} - \setminus \frac{{u}^{n + 1}}{n + 1} + C$

$= \setminus \frac{{\left(1 - x\right)}^{n + 2}}{n + 2} - \setminus \frac{{\left(1 - x\right)}^{n + 1}}{n + 1} + C$
when $n \ne - 1$ and $n \ne - 2$.

When $n = - 1$, we get:

$\setminus \int \setminus \frac{x}{1 - x} \setminus \mathrm{dx} = \setminus \int \left(\setminus \frac{1}{1 - x} - 1\right) \setminus \mathrm{dx} = - \setminus \ln | 1 - x | - x + C$

When $n = - 2$, we get:

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

$= \setminus \int \left(- \setminus \frac{1}{1 - x} + \setminus \frac{1}{{\left(1 - x\right)}^{2}}\right) \setminus \mathrm{dx} = \setminus \ln | 1 - x | + \frac{1}{1 - x} + C$