What is the integral of (16t^2+16t+4) ?

The integral of (16t^2+16t+4) is (16t^3)/3+8t^2+4t+C

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integral of (16t^2+16t+4)

\int (16 t^{2} + 16 t + 4) d t

\frac{16 t ^{3}}{3} + 8 t^{2} + 4 t + C

Solution steps

\int 16 t^{2} + 16 t + 4 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 16 t^{2} d t + \int 16 t d t + \int 4 d t

\int 16 t^{2} d t = \frac{16 t ^{3}}{3}

\int 16 t d t = 8 t^{2}

\int 4 d t = 4 t

= \frac{16 t ^{3}}{3} + 8 t^{2} + 4 t

Add a constant to the solution

= \frac{16 t ^{3}}{3} + 8 t^{2} + 4 t + C

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What is the integral of (16t^2+16t+4) ?
The integral of (16t^2+16t+4) is (16t^3)/3+8t^2+4t+C