The integral from a to b of f(x)/g(x) equals ln[(g(b)/g(a)]. Find g'(x), and explain your reasoning.

Answer 1

since ln(g(b)/g(a) = ln(g(b)) - ln(g(a))

g'(x) = 1/x

Integral of 1/x dx = ln(x)

Answer 2

I still do not understand how you found g'(x) to be 1/x. Could you please elaborate a little more on the reasoning behind this?

Answer 3

Good question. My reasoning is bogus. I was thinking ln(b)-ln(a)

We know that ∫dg/g = ln(g)

dg = g' dx

so, dg/dx = g' = f

how's that?

For example

∫(2x+3)/(x^2 + 3x - 7) dx

let g = x^2 + 3x - 7
dg = (2x+3) dx
and we have
∫dg/g = ln(g)
evaluate at a and b to get
ln(g(b)) - ln(g(a)) = ln[g(b)/g(a)]