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

\int 2 e^{- 0.8 x} d x

x y^{^{'}} + 3 x - 2 y = 0

e^{2 x} y^{^{'}} + 2 e^{2 x} y = 2 x

e^{- x^{2}} (1 - 2 x^{2})

x \to 6 + lim (x^{3} + c x)

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