What is the integral of 1/(-u+1) ?

The integral of 1/(-u+1) is -ln|-u+1|+C

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integral of 1/(-u+1)

\int \frac{1}{- u + 1} d u

- ln ∣ - u + 1 ∣ + C

Solution steps

\int \frac{1}{- u + 1} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

v = - u + 1

= - ln ∣ - u + 1 ∣

Add a constant to the solution

= - ln ∣ - u + 1 ∣ + C

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