What is the integral of (2v^3-v)/(v^4-v^2) ?

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

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integral of (2v^3-v)/(v^4-v^2)

\int \frac{2 v ^{3} - v}{v ^{4} - v ^{2}} d v

ln v^{2} - 1 + ln ∣ v ∣ - \frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣ + C

Solution steps

\int \frac{2 v ^{3} - v}{v ^{4} - v ^{2}} d v

Expand

\frac{2 v ^{3} - v}{v ^{4} - v ^{2}} : \frac{2 v ^{3}}{v ^{4} - v ^{2}} - \frac{v}{v ^{4} - v ^{2}}

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{2 v ^{3}}{v ^{4} - v ^{2}} d v - \int \frac{v}{v ^{4} - v ^{2}} d v

\int \frac{2 v ^{3}}{v ^{4} - v ^{2}} d v = ln v^{2} - 1

\int \frac{v}{v ^{4} - v ^{2}} d v = - ln ∣ v ∣ + \frac{1}{2} ln ∣ v + 1 ∣ + \frac{1}{2} ln ∣ v - 1 ∣

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

Simplify

= ln v^{2} - 1 + ln ∣ v ∣ - \frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣

Add a constant to the solution

= ln v^{2} - 1 + ln ∣ v ∣ - \frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣ + C

x y^{2} \frac{d y}{d x} = y^{3} - x^{2}

\int \frac{( e ^{2 x} )}{49 + e ^{4 x}} d x

\frac{d ^{2}}{d x ^{2}} (x^{\frac{6}{7}} + 5 x)

y^{^{'}} = 1 + a y

\frac{d}{d x} (cos (5 x^{2} - 7 x) (10 x - 7))

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