How do you integrate? I try again and again.

#int x^10sqrt(x-2)dx#

1 Answer
Jun 19, 2018

Use the substitution #x-2=u^2#.

Explanation:

Let

#I=intx^10sqrt(x-2)dx#

Apply the substitution #x-2=u^2#

#I=2int(u^2+2)^10u^2du#

Apply binomial expansion:

#I=2int{sum_(n=0)^10((10),(n))u^(2n)2^(10-n)}u^2du#

Rearrange:

#I=sum_(n=0)^10((10),(n))2^(11-n)intu^(2n+2)du#

Integrate directly:

#I=sum_(n=0)^10((10),(n))2^(11-n)/(2n+3)u^(2n+3)+C#

Reverse the substitution:

#I=sum_(n=0)^10((10),(n))2^(11-n)/(2n+3)(x-2)^((2n+3)/2)+C#