# Thread: how to integrate -2/(x(x + 2))

1. ## how to integrate -2/(x(x + 2))

how would you go about integrating $\displaystyle \int \frac{-2}{x(x+2)} dx$?

i dont even know where to start?

2. Originally Posted by renlok
how would you go about integrating $\displaystyle \int \frac{-2}{x(x+2)} dx$?

i dont even know where to start?
Partial Fractions

$\displaystyle \frac{-2}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2}$

$\displaystyle A(x+2) + Bx = -2$

Let $\displaystyle x = 0$ so that $\displaystyle 2A = -2$ hence $\displaystyle A = -1$

Let $\displaystyle x = -2$ such that $\displaystyle -2B = -2$ and $\displaystyle B = 1$

The integral therefore becomes $\displaystyle \int \frac{-2}{x(x+2)}\,dx = -\int \frac{1}{x} \,dx+ \int\frac{1}{x+2}\,dx$ which is simpler

3. Originally Posted by renlok
how would you go about integrating $\displaystyle \int \frac{-2}{x(x+2)} dx$?

i dont even know where to start?
Hi renlok,

You can write $\displaystyle \frac{-2}{x(x+2)}$ as the sum of two fractions.

You need to find their numerators.

$\displaystyle \frac{a}{x}+\frac{b}{x+2}=\frac{-2}{x(x+2)}$

$\displaystyle \frac{a(x+2)}{x(x+2)}+\frac{bx}{x(x+2)}=\frac{-2}{x(x+2)}$

$\displaystyle ax+2a+bx=-2+(0)x$

$\displaystyle 2a=-2,\ a=-1$

$\displaystyle a+b=0,\ b=-a,\ b=1$

Now you only need solve

$\displaystyle \int{\frac{-2}{x(x+2)}}dx=\int{\frac{1}{x+2}}dx-\int{\frac{1}{x}}dx$