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

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Answer 1 · 2017-02-18T22:14:30

$int 3/(x^2+2x+4) dx=sqrt(3) arctan((x+1)/sqrt(3)) + C$

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

$int 3/(x^2+2x+4) dx= int 3/((x+1)^2+3)dx = int (dx)/ (((x+1)/sqrt3)^2+1)$

Now substitute:

$t=(x+1)/sqrt(3)$

$dt = dx/sqrt(3)$

$int (dx)/ (((x+1)/sqrt3)^2+1) = sqrt(3) int (dt)/(1+t^2) = sqrt(3) arctan t + C$

and undoing the substitution:

$int 3/(x^2+2x+4) dx=sqrt(3) arctan((x+1)/sqrt(3)) + C$

How do you integrate 3/(x^2+2x+4)3/(x^2+2x+4) using partial fractions?