What is the integral of t^2-4t-12 ?

The integral of t^2-4t-12 is (t^3)/3-2t^2-12t+C

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integral of t^2-4t-12

\int t^{2} - 4 t - 12 d t

\frac{t ^{3}}{3} - 2 t^{2} - 12 t + C

Solution steps

\int t^{2} - 4 t - 12 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{2} d t - \int 4 t d t - \int 12 d t

\int t^{2} d t = \frac{t ^{3}}{3}

\int 4 t d t = 2 t^{2}

\int 12 d t = 12 t

= \frac{t ^{3}}{3} - 2 t^{2} - 12 t

Add a constant to the solution

= \frac{t ^{3}}{3} - 2 t^{2} - 12 t + C

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What is the integral of t^2-4t-12 ?
The integral of t^2-4t-12 is (t^3)/3-2t^2-12t+C