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

The integral of 1/(x(x^5-1)) is 1/5 (-ln|x^5|+ln|-1+x^5|)+C

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

\int \frac{1}{x ( x ^{5} - 1 )} d x

\frac{1}{5} (- ln x^{5} + ln - 1 + x^{5}) + C

Solution steps

\int \frac{1}{x ( x ^{5} - 1 )} d x

Apply u-substitution

: \int \frac{1}{5 ( u + 1 ) u} d u

= \int \frac{1}{5 ( 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}{5} \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}{5} \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}{5} (- \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}{5} (- ln ∣ u + 1 ∣ + ln ∣ u ∣)

Substitute back

u = - 1 + x^{5}

= \frac{1}{5} (- ln - 1 + x^{5} + 1 + ln - 1 + x^{5})

Simplify

\frac{1}{5} (- ln - 1 + x^{5} + 1 + ln - 1 + x^{5}) : \frac{1}{5} (- ln x^{5} + ln - 1 + x^{5})

= \frac{1}{5} (- ln x^{5} + ln - 1 + x^{5})

Add a constant to the solution

= \frac{1}{5} (- ln x^{5} + ln - 1 + x^{5}) + C

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