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

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Answer 1 · 2015-02-20T18:44:21

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

$\int \frac{3}{x^{2} + 4} d x = \int \frac{3}{4} \cdot \frac{1}{1 + \frac{x^{2}}{4}} d x =$ $int3/(x^2+4)dx=int3/4*1/(1+x^2/4)dx=$

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

$d x = 2 d t$ $dx=2dt$

Substituting in the integral:

$= \int \frac{3}{4} \cdot \frac{1}{1 + t^{2}} 2 d t = \frac{3}{2} arctan (t) =$ $=int3/4*1/(1+t^2)2dt=3/2arctan(t)=$

Going back to $x$ $x$ :

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

How do you integrate 3x2+43/(x^2+4)?