What is the integral from 0 to 1 of (x^6)/(x^7-1) ?

The integral from 0 to 1 of (x^6)/(x^7-1) is diverges

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integral from 0 to 1 of (x^6)/(x^7-1)

\int_{0}^{1} \frac{x ^{6}}{x ^{7} - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{6}}{x ^{7} - 1} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{7 u} d u

= \int_{- 1}^{0} \frac{1}{7 u} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{- 1}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{7} [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\int \frac{3}{1 + x ^{2}} d x

y^{^{''}} + y = 4 x sin (x)

x \to + 3 lim ((\frac{x - 1}{2 x - 4})^{\frac{1}{x - 3}})

y^{^{''}} = - 4 \cdot y

\int \frac{1}{16 e ^{- 3 x} + e ^{3 x}} d x

What is the integral from 0 to 1 of (x^6)/(x^7-1) ?
The integral from 0 to 1 of (x^6)/(x^7-1) is diverges