What is the integral of (8x^6-6x^5-2x^2-10x)/(2x) ?

The integral of (8x^6-6x^5-2x^2-10x)/(2x) is 1/2 ((4x^6)/3-(6x^5)/5-x^2-10x)+C

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integral of (8x^6-6x^5-2x^2-10x)/(2x)

\int \frac{8 x ^{6} - 6 x ^{5} - 2 x ^{2} - 10 x}{2 x} d x

\frac{1}{2} (\frac{4 x ^{6}}{3} - \frac{6 x ^{5}}{5} - x^{2} - 10 x) + C

Solution steps

\int \frac{8 x ^{6} - 6 x ^{5} - 2 x ^{2} - 10 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{8 x ^{6} - 6 x ^{5} - 2 x ^{2} - 10 x}{x} d x

Expand

\frac{8 x ^{6} - 6 x ^{5} - 2 x ^{2} - 10 x}{x} : \frac{8 x ^{6}}{x} - \frac{6 x ^{5}}{x} - \frac{2 x ^{2}}{x} - \frac{10 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{8 x ^{6}}{x} d x - \int \frac{6 x ^{5}}{x} d x - \int \frac{2 x ^{2}}{x} d x - \int \frac{10 x}{x} d x)

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

\int \frac{6 x ^{5}}{x} d x = \frac{6 x ^{5}}{5}

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

\int \frac{10 x}{x} d x = 10 x

= \frac{1}{2} (\frac{4 x ^{6}}{3} - \frac{6 x ^{5}}{5} - x^{2} - 10 x)

Add a constant to the solution

= \frac{1}{2} (\frac{4 x ^{6}}{3} - \frac{6 x ^{5}}{5} - x^{2} - 10 x) + C

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What is the integral of (8x^6-6x^5-2x^2-10x)/(2x) ?
The integral of (8x^6-6x^5-2x^2-10x)/(2x) is 1/2 ((4x^6)/3-(6x^5)/5-x^2-10x)+C