What is the integral of 3(x+1)2 ?

The integral of 3(x+1)2 is 6((x^2)/2+x)+C

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integral of 3(x+1)2

\int 3 (x + 1) 2 d x

6 (\frac{x ^{2}}{2} + x) + C

Solution steps

\int 3 (x + 1) \cdot 2 d x

Take the constant out:

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

= 3 \cdot 2 \cdot \int x + 1 d x

Simplify

= 6 \cdot \int x + 1 d x

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 x d x + \int 1 d x)

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

= 6 (\frac{x ^{2}}{2} + x)

Add a constant to the solution

= 6 (\frac{x ^{2}}{2} + x) + C