How do you use substitution to integrate $x \cdot {(1 - x)}^{n}$ $x*(1-x)^n$ ?

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Answer 1 · 2015-06-01T23:51:15

Let $u = 1 - x$ $u=1-x$ so that $d u = - d x$ $du = -dx$ and $x = 1 - u$ $x=1-u$ . 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:

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

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

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