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

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

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

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

diverges

Solution steps

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

Apply the Power Rule

: [- \frac{1}{3 x ^{3}}]_{0}^{1}

= [- \frac{1}{3 x ^{3}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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

\frac{d y}{d x} = 2 - 2 x - y + 3

\frac{\partial}{\partial x} (x^{3} \cdot y^{2})

\int_{1}^{0} (- 9 x) d x

\frac{\partial}{\partial t} (ln (x + 2 c t))

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