## Calculus: Early Transcendentals (2nd Edition)

$$\frac{3}{4}\left( {4 - {3^{2/3}}} \right)$$
\eqalign{ & \int_2^3 {\frac{x}{{\root 3 \of {{x^2} - 1} }}dx} \cr & {\text{substitute }}u = {x^2} - 1,{\text{ }}du = 2xdx \cr & {\text{express the limits in terms of }}u \cr & x = 2{\text{ implies }}u = {\left( 2 \right)^2} - 1 = 3 \cr & x = 3{\text{ implies }}u = {\left( 3 \right)^2} - 1 = 8 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_2^3 {\frac{x}{{\root 3 \of {{x^2} - 1} }}dx} = \int_3^8 {\frac{{1/2}}{{{u^{1/3}}}}du} \cr & {\text{find the antiderivative}} \cr & = \frac{1}{2}\left. {\left( {\frac{{{u^{2/3}}}}{{2/3}}} \right)} \right|_3^8 \cr & {\text{use the fundamental theorem}} \cr & = \frac{3}{4}\left( {{8^{2/3}} - {3^{2/3}}} \right) \cr & {\text{simplify}} \cr & = \frac{3}{4}\left( {4 - {3^{2/3}}} \right) \cr}