Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$ I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$
$$(x-3)=3\sec\theta$$
$$dx=3\sec\theta\tan\theta$$
$$\theta=arcsec(\frac{x-3}{3})$$
Here are my solving steps
$$\int\frac{x}{\sqrt{(x-3)^2-9}}dx = 3\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\sqrt{9(\sec^2\theta-1)}}d\theta$$
$$=\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\tan\theta}d\theta$$
$$=\int(3\sec\theta+3)\sec\theta\tan\theta$$
$$=3\int\sec^2\theta\tan\theta d\theta + 3\int\sec\theta\tan\theta d\theta$$
$$u = \sec\theta, du=\sec\theta\tan\theta d\theta$$
$$=3\int udu + 3\sec\theta$$
$$=\frac{3\sec\theta}{2}+3\sec\theta+C$$
$$=\frac{3\sec(arcsec(\frac{x-3}{3}))}{2}+3\sec(arcsec(\frac{x-3}{3}))+C$$
$$=\frac{3(\frac{x-3}{3})}{2}+3(\frac{x-3}{3})$$
$$=\frac{x-3}{2}+x-3+C$$
However, the expected answer is
$$\int\frac{x}{\sqrt{x^2-6x}}dx=\sqrt{x^2-6x}+3\ln\bigg(\frac{x-3}{3}+\frac{\sqrt{x^2-6x}}{3}\bigg)$$
What did I misunderstood?
 A: I prefer to apply hyperbolic substitution.
More precisely, if we let $x - 3 = 3\cosh(y)$, we arrive at
\begin{align*}
\int\frac{x}{\sqrt{x^{2} - 6x}}\mathrm{d}x & = \int\frac{x}{\sqrt{(x - 3)^{2} - 9}}\mathrm{d}x\\\\
& = 3\int\frac{(1 + \cosh(y))\sinh(y)}{\sqrt{\cosh^{2}(y) - 1}}\mathrm{d}y\\\\
& = 3\int\mathrm{d}y + 3\int \cosh(y)\mathrm{d}y\\\\
& = 3y + 3\sinh(y) + c\\\\
& = 3\operatorname{arccosh}\left(\frac{x - 3}{3}\right) + 3\sqrt{\left(\frac{x - 3}{3}\right)^{2} - 1} + c\\\\
& = 3\ln\left(\frac{x - 3}{3} + \frac{\sqrt{x^{2} - 6x}}{3}\right) + \sqrt{x^{2} - 6x} + c\\\\
& = 3\ln\left(x - 3 + \sqrt{x^{2} - 6x}\right) + \sqrt{x^{2} - 6x} + C
\end{align*}
which coincides with the proposed result because
\begin{align*}
\operatorname{arccosh}(z) = \ln\left(z + \sqrt{z^{2} - 1}\right)
\end{align*}
Hopefully this helps!
A: I'll give you an answer that is completely free from all this trigonometric nonsense, that I personally think is cleaner and easier. We wish to compute
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x.$$
Consider first rewriting it as
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\mathrm{d}x+3\int \frac{\mathrm{d}x}{\sqrt{x^2-6x}}\,.$$
Now for the first integral, letting $u=x^2-6x$ we get that
$$\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\mathrm{d}x=\int\frac{\mathrm{d}u}{2\sqrt{u}}=\sqrt{u}+A=\sqrt{x^2-6x}+A.$$
This takes care of the first integral. Now for the second integral, consider the Euler substitution
$$\begin{cases}
t=x+\sqrt{x^2-6x},\\
\mathrm{d}t=\frac{x+\sqrt{x^2-6x}-3}{\sqrt{x^2-6x}}\mathrm{d}x.
\end{cases}$$
This yields that
$$3\int \frac{1}{\sqrt{x^2-6x}}\,\mathrm{d}x=3\int \frac{\mathrm{d}t}{t-3}=3\ln\lvert t-3\rvert +D=3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+B.$$
Combining this the answer becomes
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\sqrt{x^2-6x}+3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+C.$$
A: $$
\begin{aligned}
\int \frac{x}{\sqrt{x^2-6 x}} d x & \int \frac{x}{x-3} d\left(\sqrt{x^2-6 x}\right) \\
= & \int\left(1+\frac{3}{x-3}\right)\left(\sqrt{x^2-6 x}\right) \\
= & \sqrt{x^2-6 x}+3 \int \frac{d\left(\sqrt{x^2-6 x}\right)}{\sqrt{\left(\sqrt{x^2-6 x}\right)^2+9}} \\
= & \sqrt{x^2-6 x}+3 \sinh^{-1} \left(\frac{\sqrt{x^2-6 x}}{3}\right)+C
\end{aligned}
$$
