What is the integral from 0 to 1 of (x^8)/(x^9-1) ?

The integral from 0 to 1 of (x^8)/(x^9-1) is diverges

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integral from 0 to 1 of (x^8)/(x^9-1)

\int_{0}^{1} \frac{x ^{8}}{x ^{9} - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{8}}{x ^{9} - 1} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{9 u} d u

= \int_{- 1}^{0} \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{- 1}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\int x (x^{2}) d x

\int_{1}^{2} - 2 x + 6 d x

\frac{d}{d x} (\frac{1}{x ^{16}})

y = - 8 x, y = 12, x = - 1, x = 1

\int \frac{( U - 2 )}{( U ^{2} - 3 U - 1 )} d U

What is the integral from 0 to 1 of (x^8)/(x^9-1) ?
The integral from 0 to 1 of (x^8)/(x^9-1) is diverges