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

The integral of (1/2 x+3)^2 is (x^3}{12}+\frac{3x^2)/2+9x+C

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integral of (1/2 x+3)^2

\int (\frac{1}{2} x + 3)^{2} d x

\frac{x ^{3}}{12} + \frac{3 x ^{2}}{2} + 9 x + C

Solution steps

\int (\frac{1}{2} x + 3)^{2} d x

Simplify

(\frac{1}{2} x + 3)^{2} : \frac{1}{4} x^{2} + 3 x + 9

= \int \frac{1}{4} x^{2} + 3 x + 9 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

= \int \frac{1}{4} x^{2} d x + \int 3 x d x + \int 9 d x

\int \frac{1}{4} x^{2} d x = \frac{x ^{3}}{12}

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

\int 9 d x = 9 x

= \frac{x ^{3}}{12} + \frac{3 x ^{2}}{2} + 9 x

Add a constant to the solution

= \frac{x ^{3}}{12} + \frac{3 x ^{2}}{2} + 9 x + C

n = 1 \sum \infty \frac{5}{2 ^{n - 1}}

\frac{\partial}{\partial x} (\frac{x - z}{y + z})

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\int x cos (π x) d x

\int (3 + e^{x} (y^{2} + 3 y + 2)) d x

What is the integral of (1/2 x+3)^2 ?
The integral of (1/2 x+3)^2 is (x^3}{12}+\frac{3x^2)/2+9x+C