What is the integral from 0 to 2 of 1/(4-2x) ?

The integral from 0 to 2 of 1/(4-2x) is diverges

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integral from 0 to 2 of 1/(4-2x)

\int_{0}^{2} \frac{1}{4 - 2 x} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{4 - 2 x} d x

Apply u-substitution

: \int_{4}^{0} - \frac{1}{2 u} d u

= \int_{4}^{0} - \frac{1}{2 u} d u

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

= - \int_{0}^{4} - \frac{1}{2 u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{0}^{4} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{2} [ln ∣ u ∣]_{0}^{4})

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{0}^{4}

Compute the boundaries:

diverges

= diverges

\int_{1}^{5} (2 + 3 x - x^{2}) d x

\frac{1}{10 x}

\int 2 e^{- t} d t

\int \frac{1}{x ^{2} + 2} d x

\frac{d y}{d x} = \frac{x - 2}{y - 4}

What is the integral from 0 to 2 of 1/(4-2x) ?
The integral from 0 to 2 of 1/(4-2x) is diverges