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

The integral of ((x^3-1))/((4x^3-x)) is 1/4 x+ln|x|-9/16 ln|2x+1|-7/16 ln|2x-1|+C

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

\int \frac{( x ^{3} - 1 )}{( 4 x ^{3} - x )} d x

\frac{1}{4} x + ln ∣ x ∣ - \frac{9}{16} ln ∣ 2 x + 1 ∣ - \frac{7}{16} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} - 1}{4 x ^{3} - x} : \frac{1}{4} + \frac{1}{x} - \frac{9}{8 ( 2 x + 1 )} - \frac{7}{8 ( 2 x - 1 )}

= \int \frac{1}{4} + \frac{1}{x} - \frac{9}{8 ( 2 x + 1 )} - \frac{7}{8 ( 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}{4} d x + \int \frac{1}{x} d x - \int \frac{9}{8 ( 2 x + 1 )} d x - \int \frac{7}{8 ( 2 x - 1 )} d x

\int \frac{1}{4} d x = \frac{1}{4} x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{9}{8 ( 2 x + 1 )} d x = \frac{9}{16} ln ∣ 2 x + 1 ∣

\int \frac{7}{8 ( 2 x - 1 )} d x = \frac{7}{16} ln ∣ 2 x - 1 ∣

= \frac{1}{4} x + ln ∣ x ∣ - \frac{9}{16} ln ∣ 2 x + 1 ∣ - \frac{7}{16} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{4} x + ln ∣ x ∣ - \frac{9}{16} ln ∣ 2 x + 1 ∣ - \frac{7}{16} ln ∣ 2 x - 1 ∣ + C

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

f (x) = \frac{e ^{x} x - e ^{x}}{5 x ^{2}}

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

\frac{3 x - 7}{x}

\frac{d y}{d x} = - 6 y x

What is the integral of ((x^3-1))/((4x^3-x)) ?
The integral of ((x^3-1))/((4x^3-x)) is 1/4 x+ln|x|-9/16 ln|2x+1|-7/16 ln|2x-1|+C