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February 18, 2026
Section 5.2 : Computing Indefinite Integrals
10. Evaluate \( \displaystyle \int{{\left( {{t^2} - 1} \right)\left( {4 + 3t} \right)\,dt}}\).
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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 2At 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.