MHB Integral with substitution method

[Yankel]

Hello

I need to solve

\[\int \sqrt{x^{3}+4}\cdot x^{5}dx\]

using the substitution method.

I did

\[u=x^{3}+4\]

but I got stuck with it.

thanks!

 

[MarkFL]

I would try integration by parts, where:

$$u=x^3\,\therefore\,du=3x^2\,dx$$

$$dv=x^2\sqrt{x^3+4}\,dx\, \therefore\,v=\frac{2}{9}\left(x^3+4 \right)^{\frac{3}{2}}$$

Can you proceed?

 

[awkward]

Your original idea of substitution will also work. Letting
$u = x^3+4$

we have

$du = 3 x^2 dx$
$x^3 = u-4$

so
$\int \sqrt{x^3+4} \; x^5 \;dx = \int \sqrt{x^3+4} \; x^3 \cdot x^2 \;dx = \frac{1}{3} \int \; \sqrt{u} (u-4) \; dx$

and you can probably take it from there...