I am having some difficulty with this problem. I am getting a finite answer but when I put the equation into wolfram alpha to check my answer it says that the integral does not converge.Here is what I have tried to do so far:
$$\int_0^5 \frac{dx}{x^2-x-2} \\ \int_0^5 \frac{dx}{x^2-x-2} = \int_0^5 \frac{1}{x^2-x-2}dx = \int_0^5 \frac{1}{(x+2)(x-1)}dx = \int_0^5 \frac{A}{(x+2)}dx + \int_0^5 \frac{B}{(x-1)}dx \\ \text{Find A, let x = -2} \\ 1 = A(x-1) + B(x+2) \\ 1 = -3A \\ A = -1/3 \\ \text{Find B, let x = 1}\\ 1 = A(x-1) + B(x+2) \\ 1 = 3B \\ B = 1/3 \\ \text{Rewrite the original integral as a product pulling the 1/3 out of the integral} \\ \frac{1}{3} [-\int_0^5 \frac{1}{x + 2} + \int_0^5 \frac{1}{x - 1}] \\ \frac{1}{3} [-\int_0^5 \frac{1}{u}du + \int_0^5 \frac{1}{w}dw] \\ \text{Change the bounds on the integral} \\ \text{u = 5 + 2 = 7, u = 0 + 2 = 2} \\ \text{w = 5 - 1 = 4, w= 0 - 1 = -1} \\ \frac{1}{3} [-\ln\lvert u\rvert \lvert_2^7 + \ln\lvert w\rvert \lvert_{-1}^{4} ] \\ \text{Re substitute our u and w values} \\ \frac{1}{3} [-\ln\lvert x + 2\rvert \lvert_2^7 + \ln\lvert x - 1\rvert \lvert_{-1}^{4} ] \\ \frac{1}{3} [-(\ln(9) - \ln(4)) + \ln(3)-\ln(2)] \\ \frac{1}{3} [-0.41] \\ -1.33$$
The key to this misconception is simply found when you try to look at the graph of the function.
Graph here.
You will note that over the region $0\le x\le5$, you cross a point where the function diverges to $\pm\infty$, which is what Wolfram|Alpha was noting.