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

The integral of (4x+8x^3-6)/(2x) is 1/3 (6x+4x^3-9ln|x|)+C

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

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

\frac{1}{3} (6 x + 4 x^{3} - 9 ln ∣ x ∣) + C

Solution steps

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

Expand

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

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

\int 4 d x = 4 x

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

Simplify

\frac{1}{2} (4 x + \frac{8 x ^{3}}{3} - 6 ln ∣ x ∣) : \frac{1}{3} (6 x + 4 x^{3} - 9 ln ∣ x ∣)

= \frac{1}{3} (6 x + 4 x^{3} - 9 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{3} (6 x + 4 x^{3} - 9 ln ∣ x ∣) + C

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What is the integral of (4x+8x^3-6)/(2x) ?
The integral of (4x+8x^3-6)/(2x) is 1/3 (6x+4x^3-9ln|x|)+C