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?
 A: 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}
A: The first method in the worksheet is
$$\int\frac{dx}{\sqrt{6x+x^2}}=\int\frac{dx}{\sqrt{(x+3)^2-3^2}}$$
Let $x+3=3\cosh u$. Then
$$\begin{align}\int\frac{dx}{\sqrt{6x+x^2}} & =\int\frac{3\sinh u\,du}{3\sqrt{\cosh^2u-1}}=\int du=u+C_1 \\
& =\cosh^{-1}\left(\frac{x+3}3\right)+C_1=\ln\left(\frac{x+3}3+\sqrt{\left(\frac{x+3}3\right)^2-1}\right)+C_1 \\
& =\ln\left(x+3+\sqrt{6x+x^2}\right)+C_2\end{align}$$
The second method reads
$$\int\frac{dx}{\sqrt{6x+x^2}}=\int\frac{2u\,du}{\sqrt{6u^2+u^4}}=2\int\frac{du}{\sqrt{6+u^2}}$$
Let $u=\sqrt6\sinh v$. Then
$$\begin{align}\int\frac{dx}{\sqrt{6x+x^2}} & =2\int\frac{\sqrt6\cosh v\,dv}{\sqrt{6+6\sinh^2v}}=2\int dv=2v+C_3\ \\
 & =2\sinh^{-1}\left(\frac{u}{\sqrt6}\right)+C_3=2\ln\left(\frac{\sqrt x}{\sqrt6}+\sqrt{\frac{x}6+1}\right)+C_3 \\
& =2\ln\left(\sqrt x+\sqrt{x+6}\right)+C_4=\ln\left(x+3+\sqrt{6x+x^2}\right)+C_5\end{align}$$
A couple of answers popped up while I was typesetting, but hopefully this will be considered a reasonable answer as well.
A: I believe there is an easier way to solve the problem than the suggested substitution.
Hint: Complete the square in the radical. $\frac {1}{\sqrt{6x+x^2}}=\frac {1}{\sqrt{(x+3)^2-9}}$
Now, let $u=x+3$. Now, we have an integral of the form 
$$\int \frac{dx}{\sqrt{u^2-9}}.$$ 
This integral easily yields to the trigonometric substitution $u=\sec t$
A: Let $u=\sqrt{x}, x=u^2, dx=2udu$ to get $\displaystyle2\int\frac{1}{\sqrt{6+u^2}}du$.
Then let $u=\sqrt{6}\tan\theta, du=\sqrt{6}\sec^{2}\theta d\theta$ to get
$\displaystyle2\int\sec\theta d\theta=2\ln|\sec\theta+\tan\theta|+C=2\ln\left(\sqrt{x+6}+\sqrt{x}\right)+C$
