What is the integral from 2 to 5 of 1/(x-1) ?

The integral from 2 to 5 of 1/(x-1) is 2ln(2)

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integral from 2 to 5 of 1/(x-1)

\int_{2}^{5} \frac{1}{x - 1} d x

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

\int_{2}^{5} \frac{1}{x - 1} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{u} d u

= \int_{1}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

\int_{0}^{1} \frac{4}{t ^{2} + 1} d t

\int_{0}^{\frac{π}{2}} cos (x) \frac{1}{2} d x

\int_{0}^{2} y 4 - y^{2} d y

\int_{0}^{4} (7 - x) x d x

\int_{0}^{1} x^{2} 4 x^{2} + 1 d x

What is the integral from 2 to 5 of 1/(x-1) ?
The integral from 2 to 5 of 1/(x-1) is 2ln(2)