How do you integrate $\frac{1}{x^{2} + x + 1}$ $1/(x^2+x+1)$ using partial fractions?

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How do you integrate $\frac{1}{x^{2} + x + 1}$ $1/(x^2+x+1)$ using partial fractions?

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Answer 1 · 2017-01-27T11:49:47

$\int \frac{d x}{x^{2} + x + 1} = \frac{2}{\sqrt{3}} arctan (\frac{2 x + 1}{\sqrt{3}}) + C$ $int (dx)/(x^2+x+1) = 2/(sqrt(3))arctan( (2x+1)/sqrt(3)) +C$

Explanation:

As the denominator does not have real roots, we can proceed in a different way than using partial fractions.

Start by completing the square in the denominator of the integrand functions:

$\frac{1}{x^{2} + x + 1} = \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{4}{{(2 x + 1)}^{2} + 3} = \frac{4}{3} \frac{1}{{(\frac{2 x + 1}{\sqrt{3}})}^{2} + 1}$ $1/(x^2+x+1) = 1/((x+1/2)^2+3/4) = 4/((2x+1)^2+3) = 4/3 1/(((2x+1)/sqrt(3))^2+1)$

Now substitute:

$t = \frac{2 x + 1}{\sqrt{3}}$ $t= (2x+1)/sqrt(3)$

$d t = \frac{2}{\sqrt{3}} d x$ $dt= 2/sqrt3dx$

$\int \frac{d x}{x^{2} + x + 1} = \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} + 1} = \frac{2}{\sqrt{3}} arctan t + C$ $int (dx)/(x^2+x+1) = 2/(sqrt(3)) int (dt)/(t^2+1) = 2/(sqrt(3))arctan t +C$

and substituting back $x$ $x$ :

$\int \frac{d x}{x^{2} + x + 1} = \frac{2}{\sqrt{3}} arctan (\frac{2 x + 1}{\sqrt{3}}) + C$ $int (dx)/(x^2+x+1) = 2/(sqrt(3))arctan( (2x+1)/sqrt(3)) +C$

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How do you integrate $\frac{1}{x^{2} + x + 1}$ $1/(x^2+x+1)$ using partial fractions?

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