What is the integral of 6(t+4)^2 ?

The integral of 6(t+4)^2 is 6((t^3)/3+4t^2+16t)+C

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integral of 6(t+4)^2

\int 6 (t + 4)^{2} d t

6 (\frac{t ^{3}}{3} + 4 t^{2} + 16 t) + C

Solution steps

\int 6 (t + 4)^{2} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int (t + 4)^{2} d t

Expand

(t + 4)^{2} : t^{2} + 8 t + 16

= 6 \cdot \int t^{2} + 8 t + 16 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

= 6 (\int t^{2} d t + \int 8 t d t + \int 16 d t)

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

\int 8 t d t = 4 t^{2}

\int 16 d t = 16 t

= 6 (\frac{t ^{3}}{3} + 4 t^{2} + 16 t)

Add a constant to the solution

= 6 (\frac{t ^{3}}{3} + 4 t^{2} + 16 t) + C

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What is the integral of 6(t+4)^2 ?
The integral of 6(t+4)^2 is 6((t^3)/3+4t^2+16t)+C