What is the integral from 0 to 1 of (81)/(x^5) ?

The integral from 0 to 1 of (81)/(x^5) is diverges

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integral from 0 to 1 of (81)/(x^5)

\int_{0}^{1} \frac{81}{x ^{5}} d x

diverges

Solution steps

\int_{0}^{1} \frac{81}{x ^{5}} d x

Take the constant out:

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

= 81 \cdot \int_{0}^{1} \frac{1}{x ^{5}} d x

Apply the Power Rule

: [- \frac{1}{4 x ^{4}}]_{0}^{1}

= 81 [- \frac{1}{4 x ^{4}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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

x \to 0 + lim (1 + x^{(\frac{1}{x})})

\int_{- 2}^{- 1} x - \frac{1}{x ^{2}} d x

y = \frac{9 ( 1 - sin ( x ) )}{2 cos ( x )}

\frac{\partial}{\partial x} (ln (x y))

What is the integral from 0 to 1 of (81)/(x^5) ?
The integral from 0 to 1 of (81)/(x^5) is diverges