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

The integral of 1/((v^2+v)(v-1)^2) is ln|v|-1/4 ln|v+1|-3/4 ln|v-1|-1/(2(v-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

ln ∣ v ∣ - \frac{1}{4} ln ∣ v + 1 ∣ - \frac{3}{4} ln ∣ v - 1 ∣ - \frac{1}{2 ( v - 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( v ^{2} + v ) ( v - 1 ) ^{2}} : \frac{1}{v} - \frac{1}{4 ( v + 1 )} - \frac{3}{4 ( v - 1 )} + \frac{1}{2 ( v - 1 ) ^{2}}

= \int \frac{1}{v} - \frac{1}{4 ( v + 1 )} - \frac{3}{4 ( v - 1 )} + \frac{1}{2 ( v - 1 ) ^{2}} d v

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}{v} d v - \int \frac{1}{4 ( v + 1 )} d v - \int \frac{3}{4 ( v - 1 )} d v + \int \frac{1}{2 ( v - 1 ) ^{2}} d v

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{4 ( v + 1 )} d v = \frac{1}{4} ln ∣ v + 1 ∣

\int \frac{3}{4 ( v - 1 )} d v = \frac{3}{4} ln ∣ v - 1 ∣

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

= ln ∣ v ∣ - \frac{1}{4} ln ∣ v + 1 ∣ - \frac{3}{4} ln ∣ v - 1 ∣ - \frac{1}{2 ( v - 1 )}

Add a constant to the solution

= ln ∣ v ∣ - \frac{1}{4} ln ∣ v + 1 ∣ - \frac{3}{4} ln ∣ v - 1 ∣ - \frac{1}{2 ( v - 1 )} + C

What is the integral of 1/((v^2+v)(v-1)^2) ?
The integral of 1/((v^2+v)(v-1)^2) is ln|v|-1/4 ln|v+1|-3/4 ln|v-1|-1/(2(v-1))+C