What is the integral from 1 to 3 of 1/(x-5) ?

The integral from 1 to 3 of 1/(x-5) is -ln(2)

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integral from 1 to 3 of 1/(x-5)

\int_{1}^{3} \frac{1}{x - 5} d x

- ln (2)

Decimal

- 0.69314 \dots

Solution steps

\int_{1}^{3} \frac{1}{x - 5} d x

Apply u-substitution

: \int_{- 4}^{- 2} \frac{1}{u} d u

= \int_{- 4}^{- 2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 4}^{- 2}

Compute the boundaries:

- ln (2)

= - ln (2)

\int_{1}^{8} \frac{9}{x} d x

\int_{0}^{7} 7 x 3 x + 1 d x

\int_{- 1}^{3} \frac{1}{( 2 x + 1 ) ^{2}} d x

\int_{0}^{1} (x^{3} - 1) e^{x} d x

\int_{4}^{16} e^{x} d x

What is the integral from 1 to 3 of 1/(x-5) ?
The integral from 1 to 3 of 1/(x-5) is -ln(2)