How do you integrate #int x^3sqrt(x^2-1)# using substitution?
2 Answers
Jul 29, 2016
Explanation:
Let us take the subst.
Now,
Jul 29, 2016
Here is a third solution.
Explanation:
Let
so that
and
The integral becomes
# = 1/2[2/5u^(5/2)+2/3u^(1/2)] +C#
# = 1/15[3u^(5/2)+5u^(3/2)]+C#
# = 1/15 u^(3/2)[3u+5]+C#
Back-substituting gets us
# = 1/15 (x^2-1)^(3/2)[3(x^2-1)+5]+C#
# = 1/15 (x^2-1)^(3/2)(3x^2+2)+C#