What is the integral from 0 to 1 of (61)/(x^5) ?

The integral from 0 to 1 of (61)/(x^5) is diverges

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integral from 0 to 1 of (61)/(x^5)

\int_{0}^{1} \frac{61}{x ^{5}} d x

diverges

Solution steps

\int_{0}^{1} \frac{61}{x ^{5}} d x

Take the constant out:

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

= 61 \cdot \int_{0}^{1} \frac{1}{x ^{5}} d x

Apply the Power Rule

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

= 61 [- \frac{1}{4 x ^{4}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{- 1}^{1} (3 x^{2} + 1)^{2} d x

\int \frac{- 4}{x ^{2}} d x

y = \frac{2 x - 3}{4 x + 1}

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

\frac{t}{1 + t ^{2}}

What is the integral from 0 to 1 of (61)/(x^5) ?
The integral from 0 to 1 of (61)/(x^5) is diverges