What is the integral from 0 to 8 of 1/(8-x) ?

The integral from 0 to 8 of 1/(8-x) is diverges

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integral from 0 to 8 of 1/(8-x)

\int_{0}^{8} \frac{1}{8 - x} d x

diverges

Solution steps

\int_{0}^{8} \frac{1}{8 - x} d x

Apply u-substitution

: \int_{8}^{0} - \frac{1}{u} d u

= \int_{8}^{0} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{8} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{0}^{8} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{8})

Simplify

= [ln ∣ u ∣]_{0}^{8}

Compute the boundaries:

diverges

= diverges

x^{2} arctan (3 x)

\frac{\partial}{\partial x} (e^{x - 1} cos (y))

\frac{d}{d x} (cos^{4} (sin^{3} (x)))

y = arccos (5 x)

\frac{d}{d x} (\frac{4}{x + 3})

What is the integral from 0 to 8 of 1/(8-x) ?
The integral from 0 to 8 of 1/(8-x) is diverges