Trouble with Indefinite Integral - $\int \frac{2x^{3}+8x}{\sqrt{(x^{2}+4)}}dx$

Question

$$\int \frac{2x^{3}+8x}{\sqrt{(x^{2}+4)}}dx$$

I'm having issues integrating this. I found a step through and I do not understand the method of substitution. Would anyone be able to give me a brief explanation how they would integrate this?

I get to this point - $2\int \frac{x^{3}}{\sqrt{(x^{2}+4)}}dx + 8\int \frac{x}{\sqrt{(x^{2}+4)}}dx$

and then I can't get any further.

Answer 1

Knowing you have found your path, and wanting to keep this out of the throngs of unanswered questions, I present:

$$ \begin{align*} \int \frac{2x^{3}+8x}{\sqrt{x^{2}+4}}dx &= \int \frac{2x\left( x^2+4 \right)\sqrt{x^{2}+4}}{x^2+4}dx \\ &=\int 2x\sqrt{x^2+4}\, dx. \end{align*} $$

Let

$$u=x^2+4.$$

Then

$$du = 2x \, dx.$$

We substitute and solve,

$$ \begin{align*} \int u^{\frac{1}{2}} \, du &= \frac{2}{3}u^{\frac{3}{2}}+c \\ &=\frac{2}{3}\left( x^2+4 \right)^{\frac{3}{2}}+c. \end{align*} $$

Answer 2

For the first one, note that $x^3 = x^2 \cdot x$. Let $t = x^2$ and then trigonometric substitution $t = 2 \sin w$

Second one is easy, substitute $u = x^2 + 4$.