If $\int \frac{\sqrt{2-x-x^2}}{x^2} dx = \frac{A\sqrt{2-x-x^2}}{x} + \frac{B}{4\sqrt{2}} \ln\left|\frac{4-x+4\sqrt{2-x-x^2}}{x}\right| - \sin^{-1}\left(\frac{2x+1}{3}\right) + c$ then $|A+B|$ is equal to ____.
Step-by-Step Solution
Key Concept: Completing the square in the radical and using substitution $x = 1/t$ transforms the integral into standard forms.
The integral $I = \int \frac{dx}{\sqrt{2-x-x^2}} - 2\int \frac{dx}{x^2\sqrt{2-x-x^2}} - \int \frac{dx}{x\sqrt{2-x-x^2}}$ is solved by completing the square and substituting $x = \frac{1}{t}$. After algebraic manipulation and integration, $I = -\sqrt{2-x-x^2} - \frac{1}{\sqrt{2}} \ln\left|\frac{(4-x)+\sqrt{2-x-x^2}}{4x}\right| - \sin^{-1}\left(\frac{2x+1}{3}\right) + c$ with $A = -1, B = 1$, giving $A + B = 0$.
Correct Answer: 8