What is the integral of (5x^4)/(x^5-1) ?

The integral of (5x^4)/(x^5-1) is ln|x^5-1|+C

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integral of (5x^4)/(x^5-1)

\int \frac{5 x ^{4}}{x ^{5} - 1} d x

ln x^{5} - 1 + C

Solution steps

\int \frac{5 x ^{4}}{x ^{5} - 1} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{4}}{x ^{5} - 1} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= 5 \cdot \int \frac{1}{5 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{5} ln ∣ u ∣

Substitute back

u = x^{5} - 1

= 5 \cdot \frac{1}{5} ln x^{5} - 1

Simplify

5 \cdot \frac{1}{5} ln x^{5} - 1 : ln x^{5} - 1

= ln x^{5} - 1

Add a constant to the solution

= ln x^{5} - 1 + C

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What is the integral of (5x^4)/(x^5-1) ?
The integral of (5x^4)/(x^5-1) is ln|x^5-1|+C