Question

Int_x^2/(1-x)^2017 =?

Answer 1

`intx^2/(1-x)^2014dx=1/(2016(1-x)^2016)-2/(2015(1-x)^2015)+1/(2014(1-x)^2014)+C`

Explanation:

`u=1-x`

`du=-dx`

`-du=dx`

`x=1-u`

`x^2=(1-u)^2=1-2u+u^2`

So, rewriting with the substitution, we get:

`-int(1-2u+u^2)/u^2017du=-int(1/u^2017-2/u^2016+1/u^2015)du=-int(u^-2017-2u^-2016+u^-2015)du=-(-1/2016u^-2016+2/2015u^-2015-1/2014u^-2014)+C=1/(2016u^2016)-2/(2015u^2015)+1/(2014u^2014)=1/(2016(1-x)^2016)-2/(2015(1-x)^2015)+1/(2014(1-x)^2014)+C`