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

[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

 

[M98Ranger]

To find \frac{dy}{dx} at x = -1, we can take the derivative of the original function y with respect to x and then substitute x = -1. We can use the quotient rule to find the derivative of y:

\frac{d}{dx}\left(\frac{x^2 - 2x - 9}{(2x-1)(x^2+3)}\right) = \frac{(2x-1)(x^2+3)\cdot (2x-2) - (x^2 -2x -9)(4x+1)}{(2x-1)^2(x^2+3)^2}

Substituting x = -1 into this expression, we get:

\frac{dy}{dx}\bigg|_{x=-1} = \frac{(-3)(2)(-4) - (-12)(-3)}{(3)^2(2)^2} = \frac{24 - 36}{36} = -\frac{1}{3}

Therefore, at x = -1, the slope of the function y is -\frac{1}{3}. This can be used to determine the behavior of the function at x = -1, such as if the function is increasing or decreasing at that point.