# Calculate $\int \frac{1}{x^2+x+1}\mathrm{d}x$

Define the integral $I$ as follow:

$$I=\int \dfrac{1}{x^2+x+1}\mathrm{d}x.$$

I do not know how to integrate it. Any suggestions please?

I tried a lot of methods:

• I substituted $x^2+x=u$.
• I modified the denominator $x^2+x+1=(x+1)^2-x$ and I substituted $x+1=u$.

I know that it is somewhat very easy to do it but I am stuck.

Thanks.

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Try going via $x^2+x+1=(x+\frac 12)^2+\frac 34$ and put the integrand in the standard form $\frac 1{y^2+a^2}$ – Mark Bennet May 19 '14 at 15:14
Remember that there is a standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial with no first-degree term. – Michael Hardy May 19 '14 at 15:41

Hint

Complete the square $x^2+x+1=\left(x+\frac12\right)^2+\frac34$ and change the variable to use $$\int\frac{du}{u^2+1}=\arctan u+C$$

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lol ... you write hint and give almost full answer. Cheers!! – Santosh Linkha May 19 '14 at 15:16
I didn't give explicitly the change of variable;-) – user63181 May 19 '14 at 15:17
Hope this gets to "nice answer"! – amWhy May 20 '14 at 11:38

$(x+1/2)^2+3/4$ substitute $u=x+1/2$

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complete the square as $x^2 + x + 1 = \left( x + \frac 1 2\right)^2 + \frac 3 4$, then use the substitution $\left( x + \frac 1 2\right) = u$ you will get $$\int \frac{1}{u^2 + \left(\frac {\sqrt 3 }{2}\right )^2}du$$ which I believe has got something to do with $\arctan(u)$

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$(x+1/2)=u$ will not give the next integrand? I am afraid. – Claude Leibovici May 19 '14 at 15:16
@MarkBennet thanks corrected. is should have been careful. – Santosh Linkha May 19 '14 at 15:18