# How do I evaluate $\int \frac{x+1}{(x^2+1) \sqrt{x^2-6x+1}} dx$?

I am trying to evaluate this indefinite integral: $$\int \frac{x+1}{(x^2+1) \sqrt{x^2-6x+1}} dx$$ What I tried

• Substitution $$u = \arctan(x)$$. However, no luck after this
• Find substitutions to convert $$x^2 - 6x + 1$$ into $$(a + bu)^2$$ or $$a - (b + cu)^2$$, however I could not find any
• Converted $$x^2 - 6x + 1$$ into $$(x-3)^2 - 8$$ and used the derivative of $$\sec^{-1}{(x)}$$ but not any luck after that as well.
• I suspect that it has a real solution, so the solution from wolfram alfa is not what I am looking for.

Any help/solutions would be very much appreciated!

• Do you have reasons to believe that there is an easy solution ?
– user65203
May 5, 2021 at 8:28
• Wolframalpha's answer May 5, 2021 at 8:28
• @DatBoi I found this in some real analysis course, so I assume that it has a real solution May 5, 2021 at 8:31
• @YvesDaoust It is intended as an exercise in introductory real analysis, so I would assume it is not too complicated May 5, 2021 at 8:32
• Evaluate the definite integral....(a bugbear of mine :) )
– user284001
May 5, 2021 at 8:33

Dividing numerator and denominator by $$x$$ and pulling out a factor of $$\sqrt{x}$$ gives

$$I = \int \frac{1+\frac{1}{(\sqrt{x})^2}}{\left(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2\right)\sqrt{\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2-4}}\cdot \frac{2dx}{2\sqrt{x}}$$

This suggests using the substitution $$\sqrt{x}-\frac{1}{\sqrt{x}} = 2\cosh t$$

which simplifies the integral to

$$I = \int \frac{dt}{2\cosh^2t+1} = \int \frac{\operatorname{sech}^2 t\:dt}{3-\tanh^2t}=\frac{1}{\sqrt{3}}\tanh^{-1}\left(\frac{\tanh t}{\sqrt{3}}\right)+C$$

Using that

$$\tanh^2t = 1-\operatorname{sech}^2t = 1-\frac{4x}{(x-1)^2} = \frac{x^2-6x+1}{(x-1)^2}$$

we get a final answer of

$$I = \frac{1}{\sqrt{3}}\tanh^{-1}\left(\frac{1}{|x-1|}\sqrt{\frac{x^2-6x+1}{3}}\right)+C$$

Hint:

You can rationalize with the substitution $$\frac{x-3}{\sqrt 2}=t+\frac1t.$$

Then decompose in simple fractions. This is a little tedious. To get the final answer, you need to inverse the substitution, by solving

$$t^2-\frac{x-3}{\sqrt 2}t+1=0.$$