What is the integral from 0 to 1 of (x^4)/(x^5-1) ?

The integral from 0 to 1 of (x^4)/(x^5-1) is diverges

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integral from 0 to 1 of (x^4)/(x^5-1)

\int_{0}^{1} \frac{x ^{4}}{x ^{5} - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{4}}{x ^{5} - 1} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{5 u} d u

= \int_{- 1}^{0} \frac{1}{5 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_{- 1}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

n = 0 \sum \infty 2 (- 5)^{n}

\frac{1}{s ^{3} - 7 s}

y^{^{'''}} + 64 y^{^{'}} = 0

f (x) = ln (100 sin^{2} (x))

\frac{d ^{2}}{d x ^{2}} (x^{\frac{8}{9}} + 6 x)

What is the integral from 0 to 1 of (x^4)/(x^5-1) ?
The integral from 0 to 1 of (x^4)/(x^5-1) is diverges