What is the integral of (3x^5-2)/(4x-4x^5) ?

The integral of (3x^5-2)/(4x-4x^5) is

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integral of (3x^5-2)/(4x-4x^5)

\int \frac{3 x ^{5} - 2}{4 x - 4 x ^{5}} d x

- \frac{3}{4} x - \frac{1}{2} ln ∣ x ∣ + \frac{1}{8} (ln x^{2} + 1 + 3 arctan (x)) + \frac{5}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{3 x ^{5} - 2}{4 x - 4 x ^{5}} d x

Take the partial fraction of

\frac{3 x ^{5} - 2}{4 x - 4 x ^{5}} : - \frac{3}{4} - \frac{1}{2 x} + \frac{2 x + 3}{8 ( x ^{2} + 1 )} + \frac{5}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )}

= \int - \frac{3}{4} - \frac{1}{2 x} + \frac{2 x + 3}{8 ( x ^{2} + 1 )} + \frac{5}{16 ( x + 1 )} - \frac{1}{16 ( 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{3}{4} d x - \int \frac{1}{2 x} d x + \int \frac{2 x + 3}{8 ( x ^{2} + 1 )} d x + \int \frac{5}{16 ( x + 1 )} d x - \int \frac{1}{16 ( x - 1 )} d x

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

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{2 x + 3}{8 ( x ^{2} + 1 )} d x = \frac{1}{8} (ln x^{2} + 1 + 3 arctan (x))

\int \frac{5}{16 ( x + 1 )} d x = \frac{5}{16} ln ∣ x + 1 ∣

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

= - \frac{3}{4} x - \frac{1}{2} ln ∣ x ∣ + \frac{1}{8} (ln x^{2} + 1 + 3 arctan (x)) + \frac{5}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{3}{4} x - \frac{1}{2} ln ∣ x ∣ + \frac{1}{8} (ln x^{2} + 1 + 3 arctan (x)) + \frac{5}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ + C

What is the integral of (3x^5-2)/(4x-4x^5) ?
The integral of (3x^5-2)/(4x-4x^5) is