# Integral of Rational Function

• Sep 29th 2010, 09:17 AM
centenial
Integral of Rational Function
I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.

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

But this can't be factored and completing the square gives $(x+\frac{1}{2})^2+\frac{3}{4}$

Which doesn't look any easier to integrate.

Could someone point me in the right direction?

Thanks (Worried)
• Sep 29th 2010, 09:23 AM
Quote:

Originally Posted by centenial
I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.

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

But this can't be factored and completing the square gives $(x+\frac{1}{2})^2+\frac{3}{4}$

Which doesn't look any easier to integrate.

Could someone point me in the right direction?

Thanks (Worried)

You missed a dx.

Completing the square is a good start, move on by substituting u=x+1/2.
• Sep 29th 2010, 09:36 AM
centenial
Oh... complete the square and then u substitution. That seems so obvious now that I can't believe I didn't see it.

So...

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

$
\int{\frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}}dx}
$

$
\int{\frac{1}{u^2 + \frac{3}{4}}du}
$

$
\int{\frac{1}{u^2 + \sqrt{\frac{3}{4}}^2}du}
$

$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2u}{\sqrt{3}}} + C
$

$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2x + 1}{\sqrt{3}}} + C
$

Is that right?
• Sep 30th 2010, 03:27 AM
Quote:

Originally Posted by centenial
Oh... complete the square and then u substitution. That seems so obvious now that I can't believe I didn't see it.

So...

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

$
\int{\frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}}dx}
$

$
\int{\frac{1}{u^2 + \frac{3}{4}}du}
$

$
\int{\frac{1}{u^2 + \sqrt{\frac{3}{4}}^2}du}
$

$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2u}{\sqrt{3}}} + C
$

$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2x + 1}{\sqrt{3}}} + C
$

Is that right?

Looks fine to me.
• Sep 30th 2010, 03:49 AM
Prove It
Quote:

Originally Posted by centenial
I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.

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

But this can't be factored and completing the square gives $(x+\frac{1}{2})^2+\frac{3}{4}$

Which doesn't look any easier to integrate.

Could someone point me in the right direction?

Thanks (Worried)

If you don't want to have to resort to two substitutions, substitute $x + 1 = \frac{\sqrt{3}}{2}\tan{\theta}$ so that $dx = \frac{\sqrt{3}}{2}\sec^2{\theta}\,d\theta$.

Then $\int{\frac{dx}{x^2 + x + 1}} = \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the ta}{\left(\frac{\sqrt{3}}{2}\tan{\theta}\right)^2 + \frac{3}{4}}}$

$= \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the ta}{\frac{3}{4}(\tan^2{\theta} + 1)}}$

$= \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the ta}{\frac{3}{4}\sec^2{\theta}}}$

$= \int{\frac{\frac{\sqrt{3}}{2}}{\frac{3}{4}}\,d\the ta}$

$= \int{\frac{2\sqrt{3}}{3}\,d\theta}$

$= \frac{2\sqrt{3}\,\theta}{3} + C$

$= \frac{2\sqrt{3}\arctan{\left[\frac{2\sqrt{3}}{3}(x+1)\right]}}{3} + C$.