What is the integral of 1/((1-w)^2)+1/(w-1) ?

The integral of 1/((1-w)^2)+1/(w-1) is -1/(w-1)+ln|w-1|+C

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integral of 1/((1-w)^2)+1/(w-1)

\int \frac{1}{( 1 - w ) ^{2}} + \frac{1}{w - 1} d w

- \frac{1}{w - 1} + ln ∣ w - 1 ∣ + C

Solution steps

\int \frac{1}{( 1 - w ) ^{2}} + \frac{1}{w - 1} d w

Apply u-substitution

: \int \frac{1}{u ^{2}} + \frac{1}{u} d u

= \int \frac{1}{u ^{2}} + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{2}} d u + \int \frac{1}{u} d u

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{u} + ln ∣ u ∣

Substitute back

u = w - 1

= - \frac{1}{w - 1} + ln ∣ w - 1 ∣

Add a constant to the solution

= - \frac{1}{w - 1} + ln ∣ w - 1 ∣ + C

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What is the integral of 1/((1-w)^2)+1/(w-1) ?
The integral of 1/((1-w)^2)+1/(w-1) is -1/(w-1)+ln|w-1|+C