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

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

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integral from-2 to 3 of 1/(c-x)

\int_{- 2}^{3} \frac{1}{c - x} d x

- ln ∣ c - 3 ∣ + ln ∣ c + 2 ∣

Solution steps

\int_{- 2}^{3} \frac{1}{c - x} d x

Apply u-substitution

: \int_{c + 2}^{c - 3} - \frac{1}{u} d u

= \int_{c + 2}^{c - 3} - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{c + 2}^{c - 3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{c + 2}^{c - 3}

Compute the boundaries:

ln ∣ c - 3 ∣ - ln ∣ c + 2 ∣

= - (ln ∣ c - 3 ∣ - ln ∣ c + 2 ∣)

Simplify

= - ln ∣ c - 3 ∣ + ln ∣ c + 2 ∣

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