integration - 2

[denian]

evaluate
\int_{1}^{2} \frac{x^2 - 2x - 9}{(2x-1)(x^2+3)} dx



<br>
i already get theanswer \searrow -1.09

then, given that


y = \frac{x^2 - 2x - 9}{(2x-1)(x^2+3)}

find \frac{dy}{dx} when x=-1

how to use the answer from previous stage to apply into this question?

[Vance]

I think they are different, so if
y=\frac{U}{V}
just
\frac{dy}{dx} = \frac{(\frac{dU}{dx}V)-(\frac{dV}{dx}U)}{V^2}

>>>>> Sorry changed plus to minus. really stupid I am