Answer
$\displaystyle \int_{3}^{6}\dfrac{x}{3\sqrt{x^2-8}}dx = \dfrac{2\sqrt{7} - 1}{3} \approx 1.4305$
Work Step by Step
To evaluate the integral $\displaystyle \int_{3}^{6}\dfrac{x}{3\sqrt{x^2-8}}dx$, let $u = x^2-8$.
Then we obtain $u = x^2-8 \longrightarrow du = 2xdx \longrightarrow \dfrac{du}{2} = xdx$
Before substituting, determine the new upper and lower limits of integration
Lower limit: When $x = 3$, $u = 3^2-8 = 1$
Upper limit: When $x = 6$, $u = 6^2-8 = 28$
$\displaystyle \int_{3}^{6}\dfrac{x}{3\sqrt{x^2-8}}dx = \displaystyle \int_{1}^{28}\dfrac{1}{3\sqrt{x^2-8}}(xdx) = \displaystyle \dfrac{1}{3}\int_{1}^{28}\dfrac{1}{\sqrt{u}}\dfrac{du}{2} = \displaystyle \dfrac{1}{3}\times\dfrac{1}{2}\int_{1}^{28}\dfrac{1}{u^{1/2}}du = \displaystyle \dfrac{1}{6}\int_{1}^{28}u^{-1/2}du = \dfrac{1}{6} \dfrac{u^{1/2}}{\frac{1}{2}} \Bigg\vert_{1}^{28} = \dfrac{2}{6} u^{1/2}\Bigg\vert_{1}^{28} = \dfrac{1}{3}\bigg[ \sqrt{28} - \sqrt{1} \bigg] = \dfrac{1}{3}\bigg[ 2\sqrt{7} - 1 \bigg] = \dfrac{2\sqrt{7} - 1}{3} \approx 1.4305$
