What is the integral from 1 to 2 of 1/(2-x) ?

The integral from 1 to 2 of 1/(2-x) is diverges

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integral from 1 to 2 of 1/(2-x)

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

diverges

Solution steps

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

Apply u-substitution

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

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{1})

Simplify

= [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{e ^{2 x + 1}}{2 x + 5}

\int_{0}^{2} x e^{- 4 x} d x

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

y^{^{'}} = (cos^{2} (t)) (cos^{2} (2 y))

(\frac{y ^{3}}{3}) + \frac{1}{4 y}

What is the integral from 1 to 2 of 1/(2-x) ?
The integral from 1 to 2 of 1/(2-x) is diverges