Integration - 2

1. May 15, 2004

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?

2. May 15, 2004

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

Last edited: May 15, 2004