<s>[INST] Integrate $\int \frac{\sqrt{x^2-8x+5}}{x}dx$ I've been trying this one for days and I can't seem to get it.
I tried:
$$\int \frac{\sqrt{x^2-8x+5}}{x}dx = \int \frac{\sqrt{\left(x-4\right)^2-11}}{x}dx$$
and set
$$x-4 = \sqrt{11}\sec\theta $$
so
$$dx=\sqrt{11}\sec\theta \tan\theta d\theta $$
and I had
$$\int \frac{\sqrt{x^2-8x+5}}{x}dx = \int \:\frac{11\sec\theta \tan^2\theta }{\sqrt{11}\sec\theta \:+\:4}d\theta $$
and I got stuck right there.
Any ideas?
Thanks!
 [/INST] Another possibility is to do an Euler substitution.
Let $$y=\sqrt{x^2-8x+5}-x$$
Then $$(y+x)^2=x^2-8x+5$$ and one solves to get 
$$x=\frac{5-y^2}{2(y+4)}$$
That implies that 
$$\sqrt{x^2-8x+5}=y+x=y+\frac{5-y^2}{2(y+4)}$$
Thus the integral reduces to a rational function. The denominator is not that bad, I get $(y+4)^2(5-y^2)$. I dont feel like calculating the rest but the remainder should be straightforward partial fraction decomposition. Good Luck ! 
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