What is the integral of ((12x^4+3x^2+6ax)/(2x)) ?

The integral of ((12x^4+3x^2+6ax)/(2x)) is 1/2 (3x^4+(3x^2)/2+6ax)+C

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integral of ((12x^4+3x^2+6ax)/(2x))

\int (\frac{12 x ^{4} + 3 x ^{2} + 6 a x}{2 x}) d x

\frac{1}{2} (3 x^{4} + \frac{3 x ^{2}}{2} + 6 a x) + C

Solution steps

\int \frac{12 x ^{4} + 3 x ^{2} + 6 a x}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{12 x ^{4} + 3 x ^{2} + 6 a x}{x} d x

Expand

\frac{12 x ^{4} + 3 x ^{2} + 6 a x}{x} : \frac{12 x ^{4}}{x} + \frac{3 x ^{2}}{x} + \frac{6 a x}{x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{12 x ^{4}}{x} d x + \int \frac{3 x ^{2}}{x} d x + \int \frac{6 a x}{x} d x)

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

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

\int \frac{6 a x}{x} d x = 6 a x

= \frac{1}{2} (3 x^{4} + \frac{3 x ^{2}}{2} + 6 a x)

Add a constant to the solution

= \frac{1}{2} (3 x^{4} + \frac{3 x ^{2}}{2} + 6 a x) + C

x \to 3 + lim (\frac{7}{x - 3})

\int \frac{2 x + 4}{x ^{2}} d x

n = 0 \sum \infty \frac{1}{( 2 ^{n} )}

P - P^{2} = \frac{d P}{d t}

\frac{\partial}{\partial x} (e^{\frac{- y}{x}})

What is the integral of ((12x^4+3x^2+6ax)/(2x)) ?
The integral of ((12x^4+3x^2+6ax)/(2x)) is 1/2 (3x^4+(3x^2)/2+6ax)+C