Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've to evaluate the integral $$\int_1^4 \frac{dx}{x^2+x+1}$$

but I can't find the answer. I checked with Wolfram Alpha but I still don't fully understand. Could you please explain the steps to me? I think I should use arctan in my answer.

share|improve this question
1  
This is not an indefinite integral, but I don't have tag privileges yet... could someone help me out? :) –  anorton Nov 21 '12 at 0:55
add comment

1 Answer

up vote 5 down vote accepted

First note that $$\dfrac1{x^2 + x + 1} = \dfrac1{(x+1/2)^2 + (\sqrt{3}/2)^2}$$ We now have that \begin{align} I = \int_1^4 \dfrac{dx}{x^2 + x + 1} = \int_1^4 \dfrac{dx}{(x+1/2)^2 + (\sqrt{3}/2)^2} \end{align} Now let $y = x+1/2$, then we get that $$I = \int_{3/2}^{9/2} \dfrac{dy}{y^2 + (\sqrt{3}/2)^2}$$ Now let $y = \sqrt{3}/2 \tan(\theta)$. We then get that $dy = \sqrt{3}/2 \sec^2(\theta) d \theta$. Hence, $$I = \int_{\theta = \arctan(\sqrt{3})}^{\theta = \arctan(3 \sqrt{3})} \dfrac{\sqrt{3}/2 \sec^2(\theta) d \theta}{3/4 \sec^2(\theta)} = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \arctan(\sqrt{3}) \right) = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \dfrac{\pi}3 \right)$$

share|improve this answer
    
That was my first step as well..... then I substituted u=x+.5 and got lost again... –  Bob Nov 21 '12 at 0:49
    
Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}u$. You will end up (apart from constant in front) integrating $\frac{1}{u^2+1}\,du$. –  André Nicolas Nov 21 '12 at 0:53
    
Oh, yeah I guess that's "easier". I first substituted $u=x+.5$ and afterwards $v=\frac{2u}{\sqrt3}$. But it appears to be quite hard... These are questions I have to solve "at home" (so that means, the easy ones) and regarding that I only had 1 hour lecture about integrals I don't think I should be capable of doing this.... Do you think I should? –  Bob Nov 21 '12 at 0:57
    
@Bob If you are familiar with the fact that $$\int \dfrac{du}{u^2+1} = \arctan(u) + c$$ then I do not think that this is hard and you should be able to do it. –  user17762 Nov 21 '12 at 0:59
    
I do know that, cause that is in my book. But ok then I need to study some more integrals :) –  Bob Nov 21 '12 at 1:01
add comment

Your Answer

 
discard

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.