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I need to calculate the integral of $1 / (2x^2+2x+1)$.

I used WolframAlpha and get this answer: $\tan^{-1}(2x+1)$ but I don't understand how to get there.

Can you help?

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    complete the square in the denominator.2012-03-15

4 Answers 4

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Well, there are a few general methods that you want to employ when you see a rational function like that. Remember to look for completing the square, inverse trigonometric substitutions, direct substitutions, integration by parts, or partial fractions. In this case, it seems like completing the square or maybe even just factoring it might work, I'll give it a shot.

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

From here I would recommend using a direct substituion of $u=x + \frac{1}{2}$ and continuing from here as suggested in the other answers.

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$2/((4x^2+4x+1)+1)$ just multiply the numerator and denominator by 2

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    @AsafKaragila, thanks for the explanation. This was one of the first times SW managed to call my attention to a list of flagged posts (+10k users have the privilege to reinforce/water down such messages to moderators), and I'm inexperienced at that. It does sound like I did the right think in directing the moderators' attention elsewhere?2012-03-15
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First note that $\displaystyle \int \frac{dy}{a^2+y^2} = \frac1a \tan^{-1}\left(\frac{y}{a} \right)$. This is obtained by substituting $y = a \tan( \theta)$ in the integrand and integrating.

For your problem, $I = \displaystyle \int \frac{dx}{2x^2 + 2x+1}$. First complete the square in the denominator i.e. rewrite $2x^2 + 2x + 1$ as $2 \left(x^2 + x + \frac12 \right) = 2 \left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)$. Hence, we get $I = \displaystyle \int \frac{dx}{2x^2 + 2x+1} = \int \frac{dx}{2 \left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)} = \frac12 \int \frac{dx}{\left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)}.$ Setting $y = x + \frac12$, we get $I = \frac12 \int \frac{dy}{y^2 + \left(\frac12 \right)^2} = \tan^{-1}(2y) = \tan^{-1}(2x+1).$

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    Thanks. That was what I had done but I made a mistake after setting $y$.2012-03-15
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Here's the method to completion. If it still doesn't make sense, post your progress and we can proceed from there.

You should complete the square on the bottom, use a u-substitution on the squared part, and use the regular arctan integral.

Does that make sense?

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    @JustinDomingue: See my answer.2012-03-15