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I need to solve the following integral

$$\int\frac{x}{\sqrt{x^2-6x}}$$

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

$$\sqrt{x^2-6x}+3ln(\frac{x-3}{3}+\frac{\sqrt{x^2-6x}}{3})$$

What did I misunderstood ?

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    $\begingroup$ When you go from line 12 to 13 you have not simplified $\tan\theta.$ That is, line 13 should read $$\int(3\sec\theta+3)\sec\theta d\theta.$$ $\endgroup$ – mfl Mar 14 '15 at 20:35
  • $\begingroup$ @mfl I was able to solve it with your correction :) Feel free to post that as an answer if you want me to accept. $\endgroup$ – student Mar 14 '15 at 20:54
  • $\begingroup$ It is not necessary. You have solved the problem by yourself. It was only a typo. Good luck. $\endgroup$ – mfl Mar 14 '15 at 20:55
  • $\begingroup$ @mfl Well, thanks a lot for your hint. $\endgroup$ – student Mar 14 '15 at 20:56
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Trough a lot of trouble I got:

$$\int \frac{x}{\sqrt{x^2-6x}}dx=\frac{x(x-6)+6\sqrt{x-6}\sqrt{x}*ln\left(\sqrt{x-6}\sqrt{x}\right)}{\sqrt{(x-6)x}}+C$$

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