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

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

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integral of (6x^2-2x-1)/(4x^3-x)

\int \frac{6 x ^{2} - 2 x - 1}{4 x ^{3} - x} d x

ln ∣ x ∣ + \frac{3}{4} ln ∣ 2 x + 1 ∣ - \frac{1}{4} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{6 x ^{2} - 2 x - 1}{4 x ^{3} - x} d x

Take the partial fraction of

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

= \int \frac{1}{x} + \frac{3}{2 ( 2 x + 1 )} - \frac{1}{2 ( 2 x - 1 )} 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

= \int \frac{1}{x} d x + \int \frac{3}{2 ( 2 x + 1 )} d x - \int \frac{1}{2 ( 2 x - 1 )} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{2 ( 2 x + 1 )} d x = \frac{3}{4} ln ∣ 2 x + 1 ∣

\int \frac{1}{2 ( 2 x - 1 )} d x = \frac{1}{4} ln ∣ 2 x - 1 ∣

= ln ∣ x ∣ + \frac{3}{4} ln ∣ 2 x + 1 ∣ - \frac{1}{4} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ + \frac{3}{4} ln ∣ 2 x + 1 ∣ - \frac{1}{4} ln ∣ 2 x - 1 ∣ + C

\frac{\partial}{\partial x} (sin (x) \cdot sin (x + y))

\int sin^{5} (8 x) d x

f (x) = 4 x^{2}

\frac{d}{d x} (\frac{x ^{2} - 1}{e ^{x}})

\frac{d}{d x} (- 9 sin (x))

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