$\displaystyle \int \frac{4dx}{x(x^2+4)}$
Answer:
$\displaystyle ln(x)-\frac{1}{2}ln(x^2-4)+c$
By partial fractions:
$\displaystyle \frac{1}{x(x^2+4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$
$\displaystyle 1 = A(x^2 + 4) + x(Bx + C)$
$\displaystyle 1 = Ax^2 + 4A + Bx^2 + Cx$
$\displaystyle 1 = (A+B)x^2 + Cx + 4A$
To get the 1, we want to eliminate the x's. One way is to set x = 0 and get $\displaystyle A = \frac{1}{4}$. The other is to get A + B = 0 and C = 0. So, $\displaystyle B = -\frac{1}{4}$, $\displaystyle C = 0$
So, we get: $\displaystyle 4\int \frac{1}{x(x^2 + 4)} \: dx \: \: = \: \: 4\int \left(\frac{1}{4x} + \frac{x}{-4(x^2+4)}\right) \: dx \: \: = \: \: \int \left(\frac{1}{x} - \frac{x}{x^2+4}\right)dx$