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

The integral from 0 to 1 of 1/(x-p) is ln|-p+1|-ln|p|

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

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

ln ∣ - p + 1 ∣ - ln ∣ p ∣

Solution steps

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

Apply u-substitution

: \int_{- p}^{- p + 1} \frac{1}{u} d u

= \int_{- p}^{- p + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- p}^{- p + 1}

Compute the boundaries:

ln ∣ - p + 1 ∣ - ln ∣ p ∣

= ln ∣ - p + 1 ∣ - ln ∣ p ∣

\int_{0}^{3} \frac{2}{3 x - 2} d x

\int_{7}^{10} (\frac{x}{2} - 3) d x

\int_{1}^{4} \frac{1}{4 - x} d x

\int_{- 5}^{5} 3 x^{2} - 6 x - 9 d x

\int_{0}^{1} \frac{13}{4} + \frac{9}{4} x d x

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