What is the integral of 1/(x(x^8-1)) ?

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

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integral of 1/(x(x^8-1))

integral

\frac{1}{x ( x ^{8} - 1 )}

\frac{1}{8} (- 8 ln (x) + ln - 1 + x^{8}) + C

Solution steps

\int \frac{1}{x ( x ^{8} - 1 )} d x

Apply u-substitution

: \int \frac{1}{8 ( u + 1 ) u} d u

= \int \frac{1}{8 ( u + 1 ) u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{( u + 1 ) u} d u

Take the partial fraction of

\frac{1}{( u + 1 ) u} : - \frac{1}{u + 1} + \frac{1}{u}

= \frac{1}{8} \cdot \int - \frac{1}{u + 1} + \frac{1}{u} d u

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}{8} (- \int \frac{1}{u + 1} d u + \int \frac{1}{u} d u)

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{8} (- ln ∣ u + 1 ∣ + ln ∣ u ∣)

Substitute back

u = - 1 + x^{8}

= \frac{1}{8} (- ln - 1 + x^{8} + 1 + ln - 1 + x^{8})

Simplify

\frac{1}{8} (- ln - 1 + x^{8} + 1 + ln - 1 + x^{8}) : \frac{1}{8} (- 8 ln (x) + ln - 1 + x^{8})

= \frac{1}{8} (- 8 ln (x) + ln - 1 + x^{8})

Add a constant to the solution

= \frac{1}{8} (- 8 ln (x) + ln - 1 + x^{8}) + C

\frac{( 8 x + 2 )}{x}

\frac{5}{x ^{4}}

\frac{( 3 x - 2 )}{( 2 x - 1 )}

x - \frac{1}{2} cos (x^{\frac{1}{2}})

- 2 cos (2 x)

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