What is the integral from 0 to 1 of (x^3)/(x^4-1) ?

The integral from 0 to 1 of (x^3)/(x^4-1) is diverges

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integral from 0 to 1 of (x^3)/(x^4-1)

\int_{0}^{1} \frac{x ^{3}}{x ^{4} - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{3}}{x ^{4} - 1} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{4 u} d u

= \int_{- 1}^{0} \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{- 1}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (9 x + 2)

y (1 - t) d t + t^{2} (1 - t) d y = 0

\frac{d}{d x} (- \frac{2 x}{( x ^{2} + 1 ) ^{2}})

f (x) = 4 x^{2} - 3 x, at x = - 1

x \to 2 lim (x^{2} + 2 x - 7)

What is the integral from 0 to 1 of (x^3)/(x^4-1) ?
The integral from 0 to 1 of (x^3)/(x^4-1) is diverges