What is the integral of 4(t+3)^2 ?

The integral of 4(t+3)^2 is 4((t^3)/3+3t^2+9t)+C

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integral of 4(t+3)^2

\int 4 (t + 3)^{2} d t

4 (\frac{t ^{3}}{3} + 3 t^{2} + 9 t) + C

Solution steps

\int 4 (t + 3)^{2} d t

Take the constant out:

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

= 4 \cdot \int (t + 3)^{2} d t

Expand

(t + 3)^{2} : t^{2} + 6 t + 9

= 4 \cdot \int t^{2} + 6 t + 9 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

= 4 (\int t^{2} d t + \int 6 t d t + \int 9 d t)

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

\int 6 t d t = 3 t^{2}

\int 9 d t = 9 t

= 4 (\frac{t ^{3}}{3} + 3 t^{2} + 9 t)

Add a constant to the solution

= 4 (\frac{t ^{3}}{3} + 3 t^{2} + 9 t) + C

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What is the integral of 4(t+3)^2 ?
The integral of 4(t+3)^2 is 4((t^3)/3+3t^2+9t)+C