Question

the integral of (x/(4x^2+9))

Answer 1

Let me give you a hint to start you off:
U = 4x^2 + 9
dU = 8x
dU/8 = x

Answer 2

this is a log function as the derivative of the denominator is 8x and the numerator is x...

so you basically have \[\frac{f'(x)}{f(x)}\]

so making an adjustment you have

\[\frac{1}{8} \times \frac{8x}{4x^2 + 9}\]

which is

\[\frac{1}{8} \int\limits \frac{8x}{4x^2 + 9} dx = \frac{1}{8} \ln(4x^2 + 9) + C\]

this can be simplified further by using log and index laws.

hope this helps

Answer 3

thank you

Answer 4

Medal? :D