integration - Integrating $frac {1}{sqrt{6x+x^2}}$ using a specified U-substitution?

I'm doing a two part homework question based on finding the integral of $\frac {1}{\sqrt{6x+x^2}}$

The first part was pretty simple, just completing the square, and recognizing it as being arcsine.

The second part is a bit tougher, since it wants the integral to be found using u-substitution, specifically having $u = \sqrt{x}$.

I've tried doing some algebra to enable such substitution, but I can't really see a way to split up the radical and make it happen.

I'm starting to think the question can't be done, and is simply a typo or something in the worksheet, since I found another more obvious error a few problems earlier.

Am I simply missing something, or am I right in assuming the worksheet is wrong?

Answer

As mentioned in the comments by Archis, the integral is $sinh^{-1} \left( \frac{u}{\sqrt{6}} \right) + C$.

$$\begin{align*} \int \frac{dx}{\sqrt{x^2 + 6x}} = \int \frac{dx}{\sqrt{(x + 3)^2 - 9}} \end{align*}$$

Let $\quad$ $u = x^{\frac{1}{2}}, \\ du = \frac{1}{2}x^{-\frac{1}{2}}dx \\ dx = 2x^{\frac{1}{2}}du \rightarrow 2udu.$

$\\$

This yields

$$\begin{align*} \int \frac{2udu}{\sqrt{(u^2 + 3)^2 - 9}} = \int \frac{2udu}{\sqrt{u^4 + 6u^2}} = \int \frac{2du}{\sqrt{u^2 + 6}}. \end{align*}$$

Recall that

$$\begin{align} \int \frac{dx}{\sqrt{x^2 + a^2}} = sinh^{-1}\left(\frac{x}{a} \right) + C. \end{align}$$

Thus,

$$\begin{align} \int \frac{2du}{\sqrt{u^2 + 6}} = 2sinh^{-1}\left(\frac{u}{\sqrt{6}} \right) + C &&\text{where $u = \sqrt{x}$} \end{align}$$