Complete the following activity :\(\displaystyle\int_0^2 \frac{dx}{4 + x - x^2} = \int_0^2 \frac{dx}{-x^2 + \square + \square}\)\(\displaystyle= \int_0^2 \frac{dx}{-x^2 + x + \frac{1}{4} - \square + 4}\)\(\displaystyle= -\int_0^2 \frac{dx}{\left(x -…

Answer: the boxes are \(x\), \(4\); \(\dfrac{1}{4}\); \(\dfrac{\sqrt{17}}{2}\).

\(\displaystyle\int_0^2 \frac{dx}{4 + x - x^2} = \int_0^2 \frac{dx}{-x^2 + x + 4} = \int_0^2 \frac{dx}{-x^2 + x + \frac{1}{4} - \frac{1}{4} + 4}\)

\(\displaystyle= \int_0^2 \frac{dx}{\frac{17}{4} - \left(x - \frac{1}{2}\right)^2} = -\int_0^2 \frac{dx}{\left(x - \frac{1}{2}\right)^2 - \left(\frac{\sqrt{17}}{2}\right)^2}\)

Using \(\displaystyle\int \frac{dx}{a^2 - X^2} = \frac{1}{2a}\log\left|\frac{a + X}{a - X}\right|\) with \(a = \frac{\sqrt{17}}{2}\), \(X = x - \frac{1}{2}\), \(\frac{1}{2a} = \frac{1}{\sqrt{17}}\):

\(\displaystyle= \frac{1}{\sqrt{17}}\left[\log\left|\frac{\sqrt{17} + 2x - 1}{\sqrt{17} - 2x + 1}\right|\right]_0^2 = \frac{1}{\sqrt{17}}\left[\log\frac{\sqrt{17} + 3}{\sqrt{17} - 3} - \log\frac{\sqrt{17} - 1}{\sqrt{17} + 1}\right]\)

\(\displaystyle= \frac{1}{\sqrt{17}}\log\frac{(\sqrt{17} + 3)(\sqrt{17} + 1)}{(\sqrt{17} - 3)(\sqrt{17} - 1)} = \frac{1}{\sqrt{17}}\log\left(\frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}}\right)\).