How do you integrate $\frac{3}{x^{2} + 2 x + 4}$ $3/(x^2+2x+4)$ using partial fractions?

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How do you integrate $\frac{3}{x^{2} + 2 x + 4}$ $3/(x^2+2x+4)$ using partial fractions?

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Answer 1 · 2017-02-18T22:14:30

$\int \frac{3}{x^{2} + 2 x + 4} d x = \sqrt{3} arctan (\frac{x + 1}{\sqrt{3}}) + C$ $int 3/(x^2+2x+4) dx=sqrt(3) arctan((x+1)/sqrt(3)) + C$

Explanation:

Rather than partial fractions I would use a substitution.
Complete the square at the denominator:

$\int \frac{3}{x^{2} + 2 x + 4} d x = \int \frac{3}{{(x + 1)}^{2} + 3} d x = \int \frac{d x}{{(\frac{x + 1}{\sqrt{3}})}^{2} + 1}$ $int 3/(x^2+2x+4) dx= int 3/((x+1)^2+3)dx = int (dx)/ (((x+1)/sqrt3)^2+1)$

Now substitute:

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

$d t = \frac{d x}{\sqrt{3}}$ $dt = dx/sqrt(3)$

$\int \frac{d x}{{(\frac{x + 1}{\sqrt{3}})}^{2} + 1} = \sqrt{3} \int \frac{d t}{1 + t^{2}} = \sqrt{3} arctan t + C$ $int (dx)/ (((x+1)/sqrt3)^2+1) = sqrt(3) int (dt)/(1+t^2) = sqrt(3) arctan t + C$

and undoing the substitution:

$\int \frac{3}{x^{2} + 2 x + 4} d x = \sqrt{3} arctan (\frac{x + 1}{\sqrt{3}}) + C$ $int 3/(x^2+2x+4) dx=sqrt(3) arctan((x+1)/sqrt(3)) + C$

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How do you integrate $\frac{3}{x^{2} + 2 x + 4}$ $3/(x^2+2x+4)$ using partial fractions?

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