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

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

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

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

\frac{1}{6} (\frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} - 8 x) + C

Solution steps

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

Simplify

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

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

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

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

\int 8 d x = 8 x

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

Add a constant to the solution

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

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