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

Question

3770 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-01T23:51:15

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#