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

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

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

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

- ln (3)

Decimal

- 1.09861 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 3}^{- 1}

Compute the boundaries:

- ln (3)

= - ln (3)

\int_{1}^{2} 4 x^{3} + 3 x^{2} + 4 x + 2 d x

\int_{0}^{2 π} x \cdot sin (n x) d x

\int_{0}^{\frac{π}{2}} 2 cos^{4} (x) d x

\int_{- r}^{r} (r^{2} - x^{2}) d x

\int_{0}^{\frac{x}{3}} sin (3 t) d t

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