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!

  • $\begingroup$ Do you have reasons to believe that there is an easy solution ? $\endgroup$
    – user65203
    May 5, 2021 at 8:28
  • $\begingroup$ Wolframalpha's answer $\endgroup$
    – DatBoi
    May 5, 2021 at 8:28
  • $\begingroup$ @DatBoi I found this in some real analysis course, so I assume that it has a real solution $\endgroup$
    – Stamatis
    May 5, 2021 at 8:31
  • $\begingroup$ @YvesDaoust It is intended as an exercise in introductory real analysis, so I would assume it is not too complicated $\endgroup$
    – Stamatis
    May 5, 2021 at 8:32
  • 1
    $\begingroup$ Evaluate the definite integral....(a bugbear of mine :) ) $\endgroup$
    – user284001
    May 5, 2021 at 8:33

3 Answers 3


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

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



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

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


With $x=y+3$ $$I=\int \frac{x+1}{(x^2+1) \sqrt{x^2-6x+1}} dx =\int \frac{y+4}{(y^2+6y+10) \sqrt{y^2-8}} dy $$ Then, substitute $t=\frac{y-\sqrt8}{\sqrt{y^2-8}}$, or $y=\frac{\sqrt8(1+t^2)}{1-t^2}$ \begin{align} I& =\frac{2\sqrt2}3\int \frac{\frac{\sqrt2+1}{t^2}-(\sqrt2-1)}{\frac{(\sqrt2+1)^2}{t^2}+(\sqrt2-1)^2t^2 -\frac23}dt\\ &= \frac1{\sqrt3}\coth^{-1}\frac{\sqrt3 [(\sqrt2-1)t-\frac{\sqrt2+1}t]}{2\sqrt2}+C\\ &= \frac1{\sqrt3}\coth^{-1}\frac{\sqrt3 (x-1)}{\sqrt{x^2-6x+1}}+C\\ \end{align}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .