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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!

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  • $\begingroup$ What approaches have you tried? $\endgroup$ – Brian Tung Oct 7 '16 at 19:56
  • $\begingroup$ I've tried Trigonometric Substitution, but it became more complicated. $\endgroup$ – glider Oct 7 '16 at 19:59
  • $\begingroup$ What trigonometric substitution? It looks like a homework to me. $\endgroup$ – user261263 Oct 7 '16 at 20:01
  • $\begingroup$ Sorry, I tried to learn how to write math in the comment. I don't know how to do it. $\endgroup$ – glider Oct 7 '16 at 20:03
  • $\begingroup$ Do it in the OP $\endgroup$ – user261263 Oct 7 '16 at 20:03
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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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  • $\begingroup$ Thank you for the solution. I didn't know about the Euler substitution method. However, I think this problem may have another solution because I get this problem from a book, and that book doesn't have the Euler substitution method. Do you have any other idea? $\endgroup$ – glider Oct 8 '16 at 3:49
  • $\begingroup$ @glider There are multiple ways to solve many of these integrals. Here the expression under the root factors, so for example if you have an integral with $\sqrt{x^2-1}$ you could write it as $\sqrt{\frac{x+1}{x-1}}(x-1)$ then your substitution is $y=\sqrt{\frac{x+1}{x-1}}$. $\endgroup$ – Rene Schipperus Oct 8 '16 at 4:31
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Trig substitution complicates it. A hyperbolic one works better. I tried $x-4=cosh(\theta)$ and then used the identity $cosh(\theta)^2-sinh(\theta)^2)=1$ to get it into a form which is straightforward. $11\int{sinh(\theta)^2.d\theta\over 4+\sqrt{11}cosh(\theta)}$

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    $\begingroup$ It should be $x-4=\sqrt{11}\cosh t$. $\endgroup$ – egreg Oct 7 '16 at 21:24
  • $\begingroup$ @susan Can you hint me how to solve the integral $11\int{sinh(\theta)^2.d\theta\over 4+\sqrt{11}cosh(\theta)}$ ? I cannot figure it out. I tried u-sub, but it didn't work. $\endgroup$ – glider Oct 8 '16 at 3:44

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