How do you integrate int (x^2-1)/sqrt(2x-1)dx?

2 Answers
Noah G
Jan 6, 2017

The answer is 1/20(2x - 1)^(5/2) + 1/6(2x - 1)^(3/2) - 3/4(2x - 1)^(1/2) + C

Explanation:

Let u = 2x - 1. Then du = 2dx and dx = 1/2du.

int(x^2 - 1)/sqrt(u) * 1/2du

We want to get rid of the x's. x = (u + 1)/2, so x^2 = ((u + 1)/2)^2

int(((u + 1)/2)^2 - 1)/sqrt(u) * 1/2du

Expand:

1/2int(((u^2 + 2u + 1)/4) - 1)/sqrt(u)du

1/2int(u^2 + 2u + 1 - 4)/(4sqrt(u))du

1/8int(u^2 + 2u - 3)/sqrt(u)

1/8int(u^2/sqrt(u) + (2u)/sqrt(u) - 3/sqrt(u))du

1/8int(u^(3/2) + 2u^(1/2) - 3u^(-1/2))du

1/8(2/5u^(5/2) + 4/3u^(3/2) - 6u^(1/2))

1/20u^(5/2) + 1/6u^(3/2) - 3/4u^(1/2)

1/20(2x - 1)^(5/2) + 1/6(2x - 1)^(3/2) - 3/4(2x - 1)^(1/2) + C

Hopefully this helps!

Cem Sentin
Nov 15, 2017

int (x^2-1)/sqrt(2x-1)*dx

=1/4*int (4x^2-4)/sqrt(2x-1)*dx

=1/4*int ((2x)^2-4)/sqrt(2x-1)*dx

After using u=sqrt(2x-1), 2x=u^2+1, 2dx=2u*du or dx=u*du transforms, this integral became

1/4*int ((2x)^2-4)/sqrt(2x-1)*dx

=1/4*int ((u^2+1)^2-4)/u*(u*du)

=1/4*int (u^4+2u^2-3)*du

=1/4*(u^5/5+(2u^3)/3-3u)+C

=u^5/20+u^3/6-(3u)/4+C

=1/20*(2x-1)^(5/2)+1/6*(2x-1)^(3/2)-3/4*sqrt(2x-1)+C

Related questions
See all questions in Integration by Substitution
Impact of this question
22884 views around the world
You can reuse this answer
Creative Commons License