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

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

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

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

- ln (2)

Decimal Notation

- 0.69314 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 2}^{- 1}

Compute the boundaries:

- ln (2)

= - ln (2)

x \to 2 lim (x^{2} x)

f (x) = \frac{x}{( 5 x - 9 ) ^{5}}

\int \frac{x}{4 x ^{2} + 2 x + 5} d x

\int - 2 x^{9} d x

2 x d x

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