## Calculus: Early Transcendentals (2nd Edition)

$$\frac{{8x\sqrt x }}{3} - 8\sqrt x + C$$
\eqalign{ & \int {\left( {4\sqrt x - \frac{4}{{\sqrt x }}} \right)} dx \cr & {\text{use the radical property }}\sqrt x = {x^{1/2}} \cr & = \int {\left( {4{x^{1/2}} - \frac{4}{{{x^{1/2}}}}} \right)} dx \cr & = \int {\left( {4{x^{1/2}} - 4{x^{ - 1/2}}} \right)} dx \cr & {\text{by the power rule for indefinite integrals}} \cr & = \frac{{4{x^{1/2 + 1}}}}{{1/2 + 1}} - \frac{{4{x^{ - 1/2 + 1}}}}{{ - 1/2 + 1}} + C \cr & = \frac{{4{x^{3/2}}}}{{3/2}} - \frac{{4{x^{1/2}}}}{{1/2}} + C \cr & = \frac{{8{x^{3/2}}}}{3} - 8{x^{1/2}} + C \cr & = \frac{{8x\sqrt x }}{3} - 8\sqrt x + C \cr & \cr & \cr & {\text{check the antiderivative by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {\frac{{8{x^{3/2}}}}{3} - 8{x^{1/2}} + C} \right) \cr & {\text{ = }}\frac{8}{3}\left( {\frac{3}{2}} \right){x^{1/2}} - 8\left( {\frac{1}{2}} \right){x^{ - 1/2}} + 0 \cr & {\text{simplify}} \cr & {\text{ = }}4{x^{1/2}} - 4{x^{ - 1/2}} \cr & {\text{ = }}4\sqrt x - \frac{4}{{\sqrt x }} \cr}