What is the integral of (8x^3-4x^2-3x)/(6x) ?

The integral of (8x^3-4x^2-3x)/(6x) is 1/6 ((8x^3)/3-2x^2-3x)+C

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integral of (8x^3-4x^2-3x)/(6x)

\int \frac{8 x ^{3} - 4 x ^{2} - 3 x}{6 x} d x

\frac{1}{6} (\frac{8 x ^{3}}{3} - 2 x^{2} - 3 x) + C

Solution steps

\int \frac{8 x ^{3} - 4 x ^{2} - 3 x}{6 x} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{8 x ^{3} - 4 x ^{2} - 3 x}{x} d x

Simplify

\frac{8 x ^{3} - 4 x ^{2} - 3 x}{x} : 8 x^{2} - 4 x - 3

= \frac{1}{6} \cdot \int 8 x^{2} - 4 x - 3 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

= \frac{1}{6} (\int 8 x^{2} d x - \int 4 x d x - \int 3 d x)

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 3 d x = 3 x

= \frac{1}{6} (\frac{8 x ^{3}}{3} - 2 x^{2} - 3 x)

Add a constant to the solution

= \frac{1}{6} (\frac{8 x ^{3}}{3} - 2 x^{2} - 3 x) + C

\frac{d}{d x} (\frac{x cos ( π x ) - 2}{x - 2})

\frac{\partial ^{2}}{\partial y \partial x} (x y e^{- 6 y})

y^{^{'}} - y = 7

\frac{\partial}{\partial u} (sin (u))

What is the integral of (8x^3-4x^2-3x)/(6x) ?
The integral of (8x^3-4x^2-3x)/(6x) is 1/6 ((8x^3)/3-2x^2-3x)+C