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

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

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

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

\frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} - 2 ln ∣ x ∣ + C

Solution steps

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

Simplify

\frac{2 x ^{6} + 8 x ^{4} - 4 x}{x ^{2}} : \frac{2 ( x ^{5} + 4 x ^{3} - 2 )}{x}

= \frac{1}{2} \cdot \int \frac{2 ( x ^{5} + 4 x ^{3} - 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 2 \cdot \int \frac{x ^{5} + 4 x ^{3} - 2}{x} d x

Expand

\frac{x ^{5} + 4 x ^{3} - 2}{x} : x^{4} + 4 x^{2} - \frac{2}{x}

= \frac{1}{2} \cdot 2 \cdot \int x^{4} + 4 x^{2} - \frac{2}{x} 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}{2} \cdot 2 (\int x^{4} d x + \int 4 x^{2} d x - \int \frac{2}{x} d x)

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{1}{2} \cdot 2 (\frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} - 2 ln ∣ x ∣)

Simplify

\frac{1}{2} \cdot 2 (\frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} - 2 ln ∣ x ∣) : \frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} - 2 ln ∣ x ∣

= \frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} - 2 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} - 2 ln ∣ x ∣ + C

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