Section 5.2 : Computing Indefinite Integrals
10. Evaluate \( \displaystyle \int{{\left( {{t^2} - 1} \right)\left( {4 + 3t} \right)\,dt}}\).
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Hint : Remember that there is no “Product Rule” for integrals and so we’ll need to eliminate the product before integrating.
Start Solution
Since there is no “Product Rule” for integrals we’ll need to multiply the terms out prior to integration.
\[\int{{\left( {{t^2} - 1} \right)\left( {4 + 3t} \right)\,dt}} = \int{{3{t^3} + 4{t^2} - 3t - 4\,dt}}\]Show Step 2
At this point there really isn’t too much to do other than to evaluate the integral.
\[\int{{\left( {{t^2} - 1} \right)\left( {4 + 3t} \right)\,dt}} = \int{{3{t^3} + 4{t^2} - 3t - 4\,dt}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{4}{t^4} + \frac{4}{3}{t^3} - \frac{3}{2}{t^2} - 4t + c}}\]Don’t forget to add on the “+c” since we know that we are asking what function did we differentiate to get the integrand and the derivative of a constant is zero and so we do need to add that onto the answer.