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

Question

How do I find the partial fraction decomposition of $\frac{2 x}{(x + 3) (3 x + 1)}$ $(2x)/((x+3)(3x+1))$ ?
How do I find the partial fraction decomposition of $\frac{1}{x^{3} + 2 x^{2} + x}$ $(1)/(x^3+2x^2+x$ ?
How do I find the partial fraction decomposition of $\frac{x^{4} + 1}{x^{5} + 4 x^{3}}$ $(x^4+1)/(x^5+4x^3)$ ?
How do I find the partial fraction decomposition of $\frac{x^{4}}{x^{4} - 1}$ $(x^4)/(x^4-1)$ ?
How do I find the partial fraction decomposition of $\frac{t^{4} + t^{2} + 1}{(t^{2} + 1) {(t^{2} + 4)}^{2}}$ $(t^4+t^2+1)/((t^2+1)(t^2+4)^2)$ ?
How do I find the integral $\int \frac{t^{2}}{t + 4} d t$ $intt^2/(t+4)dt$ ?
How do I find the integral $\int \frac{x - 9}{(x + 5) (x - 2)} d x$ $int(x-9)/((x+5)(x-2))dx$ ?
How do I find the integral $\int \frac{1}{(w - 4) (w + 1)} d w$ $int1/((w-4)(w+1))dw$ ?
How do I find the integral $\int \frac{d x}{x^{2} {(x - 1)}^{2}}$ $intdx/(x^2(x-1)^2)$ ?
How do I find the integral $\int \frac{x^{3} + 4}{x^{2} + 4} d x$ $int(x^3+4)/(x^2+4)dx$ ?

3990 views around the world

You can reuse this answer
Creative Commons License

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 3/(x^2+4)3x2+43/(x^2+4)?