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

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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

\int 9 z^{3} e^{z} d z

f (x) = \frac{x ^{2} - 1}{( x ^{2} + x + 1 )}

\int ln (5)

y^{^{''}} + 16 y = 3 sin (5 t), y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{1 - 6 x}{3 + x})

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