What is the integral from 1 to 2 of 1/(3-z) ?

The integral from 1 to 2 of 1/(3-z) is ln(2)

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integral from 1 to 2 of 1/(3-z)

\int_{1}^{2} \frac{1}{3 - z} d z

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{1}^{2} \frac{1}{3 - z} d z

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{1}^{2})

Simplify

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\int (\frac{x ^{7} - 4 x ^{5} + 11}{x ^{5}}) d x

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

x \to 1 lim ((1 - 2 x)^{\frac{1}{x}})

f (x) = \frac{t ^{3} + 3 t}{t ^{3 - 4 t + 3}}

\frac{\partial}{\partial x} (3 sec^{2} (2 x + 3 y))

What is the integral from 1 to 2 of 1/(3-z) ?
The integral from 1 to 2 of 1/(3-z) is ln(2)