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

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

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

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

ln (2) - ln (3)

Decimal

- 0.40546 \dots

Solution steps

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

Apply u-substitution

: \int_{- 3}^{- 2} \frac{1}{u} d u

= \int_{- 3}^{- 2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 3}^{- 2}

Compute the boundaries:

ln (2) - ln (3)

= ln (2) - ln (3)

\int_{2}^{\infty} \frac{3}{x ^{4}} d x

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

\int_{1}^{e} \frac{( t ^{6} + 1 )}{t} d t

\int_{1}^{2} (6 - 5 x)^{4} d x

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