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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int 8 d x = 8 x

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

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} (t)

\int \frac{e ^{x}}{e ^{2 x} + 1} d x

\frac{d ^{2}}{d x ^{2}} (sin (5) x^{5})

(1.3) (3.5)

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