How do you integrate (x^3)((x^2 + 4)^(1/2)) dx?

1 Answer
May 6, 2016

=(1/15)(x^2+4)^(3/2)(3x^2+7)+C

Explanation:

Use the substitution x^2+4=t^2#,

Then 2x dx =2t dt. So, dx=(t/x)dt=t/sqrt((t^2-4)) dt

Now, int x^3sqrt(x^2+4) dx =int (t^2-4)^(3/2)t^2/sqrt(t^2-4)dt

=int(t^2-4)t^2dt=int(t^4-4t^2)dt

=t^5/5-t^3/3+ C

=t^3/15(3t^2-5)+C

=(1/15)(x^2+4)^(3/2)(3(x^2+4)-5)+C

=(1/15)(x^2+4)^(3/2)(3x^2+7)+C