What is the integral from 0 to 1 of (41)/(x^5) ?

The integral from 0 to 1 of (41)/(x^5) is diverges

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integral from 0 to 1 of (41)/(x^5)

\int_{0}^{1} \frac{41}{x ^{5}} d x

diverges

Solution steps

\int_{0}^{1} \frac{41}{x ^{5}} d x

Take the constant out:

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

= 41 \cdot \int_{0}^{1} \frac{1}{x ^{5}} d x

Apply the Power Rule

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

= 41 [- \frac{1}{4 x ^{4}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (\frac{x}{x - 37})

f (x) = sin (3 x^{2})

\int \frac{1}{20} d x

f (x) = e^{x^{2}}

\int - sin^{3} (x) d x

What is the integral from 0 to 1 of (41)/(x^5) ?
The integral from 0 to 1 of (41)/(x^5) is diverges