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

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

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integral of v/((v-1)^2)

integral

\frac{v}{( v - 1 ) ^{2}}

ln ∣ v - 1 ∣ - \frac{1}{v - 1} + C

Solution steps

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

Apply u-substitution

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ - \frac{1}{u}

Substitute back

u = v - 1

= ln ∣ v - 1 ∣ - \frac{1}{v - 1}

Add a constant to the solution

= ln ∣ v - 1 ∣ - \frac{1}{v - 1} + C

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