What is the integral of (3x^5+1)/(3x-3x^5) ?

The integral of (3x^5+1)/(3x-3x^5) is

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integral of (3x^5+1)/(3x-3x^5)

\int \frac{3 x ^{5} + 1}{3 x - 3 x ^{5}} d x

- x + \frac{1}{3} ln ∣ x ∣ + \frac{1}{6} (- \frac{1}{2} ln x^{2} + 1 + 3 arctan (x)) + \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{3 x ^{5} + 1}{3 x - 3 x ^{5}} d x

Take the partial fraction of

\frac{3 x ^{5} + 1}{3 x - 3 x ^{5}} : - 1 + \frac{1}{3 x} + \frac{- x + 3}{6 ( x ^{2} + 1 )} + \frac{1}{6 ( x + 1 )} - \frac{1}{3 ( x - 1 )}

= \int - 1 + \frac{1}{3 x} + \frac{- x + 3}{6 ( x ^{2} + 1 )} + \frac{1}{6 ( x + 1 )} - \frac{1}{3 ( 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 1 d x + \int \frac{1}{3 x} d x + \int \frac{- x + 3}{6 ( x ^{2} + 1 )} d x + \int \frac{1}{6 ( x + 1 )} d x - \int \frac{1}{3 ( x - 1 )} d x

\int 1 d x = x

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

\int \frac{- x + 3}{6 ( x ^{2} + 1 )} d x = \frac{1}{6} (- \frac{1}{2} ln x^{2} + 1 + 3 arctan (x))

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

ln (1 - x^{2})

\int (sin (2 x) + cos (2 x))^{2} d x

x^{- \frac{1}{2}}

y = 3 - x, y = 2 x^{2}, [0, 1]

f (x) = 2 x^{2} + 3 x, (- 2, 2)

What is the integral of (3x^5+1)/(3x-3x^5) ?
The integral of (3x^5+1)/(3x-3x^5) is