<s>[INST] 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?
 [/INST] 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
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\sqrt{x^2-6x}+3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+C.$$
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