How do you find the integral of $(x^3)((x^2 + 4)^(1/2)) dx$ ?

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Answer 1 · 2015-10-10T02:12:26

Let $u=x^2+4$ .

$int x^3(x^2 + 4)^(1/2) dx = intx^2(x^2+4)^(1/2)xdx$

Let $u=x^2+4$ , so that $du=2xdx$ and $x^2=u-4$ .

After substitution, the integral becomes:

$1/2 int (u-4)u^(1/2)du$

Distribute the $u^(1/2)$ and finish the integral.

How do you find the integral of (x^3)((x^2 + 4)^(1/2)) dx(x^3)((x^2 + 4)^(1/2)) dx?