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

The integral from 1 to 5 of 1/(x-p) is ln|-p+5|-ln|-p+1|

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

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

ln ∣ - p + 5 ∣ - ln ∣ - p + 1 ∣

Solution steps

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

Apply u-substitution

= \int_{- p + 1}^{- p + 5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- p + 1}^{- p + 5}

Compute the boundaries:

ln ∣ - p + 5 ∣ - ln ∣ - p + 1 ∣

= ln ∣ - p + 5 ∣ - ln ∣ - p + 1 ∣

\int_{0}^{x} x^{5} e^{- x} d x

\int_{2}^{4} e^{- x} (x^{2} + 5) d x

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

\int_{0}^{n} x d x

\int_{0}^{2} 4 t^{2} + 4 d t

What is the integral from 1 to 5 of 1/(x-p) ?
The integral from 1 to 5 of 1/(x-p) is ln|-p+5|-ln|-p+1|