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