How do you integrate #x^3 sqrt(x^2 + 1) dx#?

1 Answer
Andrea S.
Jan 26, 2017

#int x^3sqrt(1+x^2)dx = 1/15(3x^2-2)(1+x^2)sqrt(1+x^2) +C#

Explanation:

Substitute #t=1+x^2#, #dt = 2xdx#:

#int x^3sqrt(1+x^2)dx = 1/2 int x^2 sqrt(1+x^2) (2xdx) = 1/2 int (t-1)sqrtt dt#

Now solve the integral in #t#:

#1/2 int (t-1)sqrtt dt = 1/2 int tsqrttdt -1/2 int sqrttdt = 1/2 int t^(3/2)dt -1/2 int t^(1/2)dt = 1/2 t^(5/2)/(5/2) -1/2 t^(3/2)/(3/2) +C = 1/5 t^(5/2) -1/3t^(3/2)+C = 1/15t^(3/2) (3t-5)+C#

Substituting back #x#:

#int x^3sqrt(1+x^2)dx = 1/15(1+x^2)^(3/2) (3 (1+x^2)-5)+C#

#int x^3sqrt(1+x^2)dx = 1/15(1+x^2)sqrt(1+x^2) (3x^2-2)+C#

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