## Calculus: Early Transcendentals (2nd Edition)

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