Question

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

Answer 1

Let `u=1-x` so that `du = -dx` and `x=1-u`. Then

`\int x(1-x)^n\ dx=\int (u-1)u^{n}\ du=\int (u^{n+1}-u^{n})\ du`

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

`=\frac{(1-x)^{n+2}}{n+2}-\frac{(1-x)^{n+1}}{n+1}+C`
when `n!=-1` and `n!=-2`.

When `n=-1`, we get:

`\int \frac{x}{1-x}\ dx=\int (\frac{1}{1-x}-1)\ dx=-\ln|1-x|-x+C`

When `n=-2`, we get:

`\int \frac{x}{(1-x)^2}\ dx`

`=\int (-\frac{1}{1-x}+\frac{1}{(1-x)^2})\ dx=\ln|1-x|+1/(1-x)+C`