Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could someone help me with finding this integral

$$\int \frac{dx}{x\sqrt{1 + x + x^2}}$$

or give a hint on how to solve it.

Thanks in advance

share|cite|improve this question
Wolfram|Alpha will show you steps for the solution if you click on "Show Steps", but one might hope that there would be an easier way than what it shows... – joriki Apr 9 '12 at 23:00
yep, I know but as you said I was hoping for a batter solution and seems like i got it :), thanks Américo Tavares – Dave Apr 9 '12 at 23:14
Once again: "Solving" is the wrong word. One solves equations; one solves problems. One evaluates expressions. – Michael Hardy Apr 10 '12 at 0:03
He's created a new problem in that he doesn't know how to evaluate this integral, and is asking for help in solving that. – Waffle Sep 23 '14 at 2:40
up vote 13 down vote accepted

Since the integrand is a quadratic irrational function of the type $R(x,\sqrt{1+x+x^{2}})$, you may use the Euler substitution $\sqrt{1+x+x^{2}}=x+t$. You get

$$\begin{eqnarray*} \int \frac{dx}{x\sqrt{1+x+x^{2}}} &=&\int \frac{2}{t^{2}-1}\,dt \\ &=&-2\operatorname{arctanh}t+C \\ &=&-2\operatorname{arctanh}\left( \sqrt{1+x+x^{2}}-x\right)+C. \end{eqnarray*}$$

share|cite|improve this answer
I have just solved this on paper step by step, and I wonder shouldn't it be $ln\frac{x - a}{x + a}$ instead of arctg as it's (t^2 - 1) not plus 1 – Dave Apr 9 '12 at 23:22
@Dave: You can rewrite the integral by using the identity $\operatorname{arctanh}t=\frac{1}{2}\ln \left( t+1\right) -\frac{1}{2}\ln \left( 1-t\right) .$ – Américo Tavares Apr 9 '12 at 23:28

Make the substitution $x = \frac{1}{t}$ and this reduces to finding

$$\int \frac{\text{d}t}{\sqrt{t^2 + t + 1}}$$

which can easily be reduced to finding the standard integral:

$$ \int \frac{\text{d}z}{\sqrt{z^2 + 1}} = \sinh^{-1}(z) + C$$

This substitution can be used for finding

$$\int \frac{\text{d}x}{x\sqrt{P(x)}}$$

where $P(x)$ is a quadratic polynomial in $x$.

share|cite|improve this answer

$\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{x\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}dx$.$\;\;$ Now let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$, $dx=\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta$

to get $\displaystyle\int\frac{1}{(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2})(\frac{\sqrt{3}}{2}\sec\theta)}\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta=\int\frac{\sec\theta}{\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}}d\theta=\int\frac{1}{\frac{\sqrt{3}}{2}\sin\theta-\frac{1}{2}\cos\theta}d\theta$




share|cite|improve this answer

Yes. The substitution $x = \frac{1}{t}$ works. But check for the minus sign.

It does reduce to integral of $\frac{-dt}{\sqrt{t^2+t+1}}$, which can be reduced further to integral of $\frac{-dz}{z^2 + \frac{\sqrt3}{2}}$

Regards, Prakash

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.