# How do you integrate 3/(x^2+4)?

Feb 20, 2015

I would start by factorizing $4$ from the denominator:

$\int \frac{3}{{x}^{2} + 4} \mathrm{dx} = \int \frac{3}{4} \cdot \frac{1}{1 + {x}^{2} / 4} \mathrm{dx} =$

Now I set:
${x}^{2} / 4 = {t}^{2}$ and so $x = 2 t$ giving:

$\mathrm{dx} = 2 \mathrm{dt}$

Substituting in the integral:

$= \int \frac{3}{4} \cdot \frac{1}{1 + {t}^{2}} 2 \mathrm{dt} = \frac{3}{2} \arctan \left(t\right) =$

Going back to $x$:

$= \frac{3}{2} \arctan \left(\frac{x}{2}\right) + c$