Question

Find the integral of 1/x(x^2+4).....any ideas?

Answer 1

split it up using partial fractions
\[\frac{1}{x(x^2+4)} = \frac{A}{x} + \frac{Bx}{x^2+4}\]
\[\rightarrow A(x^2 +4) +Bx^2 = 1\]
A+B = 0
4A = 1 ----> A = 1/4
---> B = -1/4

Now the integral becomes
\[\frac{1}{4}\int\limits \frac{dx}{x} - \frac{1}{4} \int\limits \frac{x}{x^2 +4} dx\]

Answer 2

Thanks, I appreciate the start.

Answer 3

no problem