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

The integral of (2-v)/(v^2+2v-3) is 1/4 ln|v-1|-5/4 ln|v+3|+C

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

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

\frac{1}{4} ln ∣ v - 1 ∣ - \frac{5}{4} ln ∣ v + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 - v}{v ^{2} + 2 v - 3} : \frac{1}{4 ( v - 1 )} - \frac{5}{4 ( v + 3 )}

= \int \frac{1}{4 ( v - 1 )} - \frac{5}{4 ( v + 3 )} 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}{4 ( v - 1 )} d v - \int \frac{5}{4 ( v + 3 )} d v

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

\int \frac{5}{4 ( v + 3 )} d v = \frac{5}{4} ln ∣ v + 3 ∣

= \frac{1}{4} ln ∣ v - 1 ∣ - \frac{5}{4} ln ∣ v + 3 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ v - 1 ∣ - \frac{5}{4} ln ∣ v + 3 ∣ + C

\frac{d y}{d x} + (\frac{2}{x}) y = 8 x

\frac{\partial}{\partial x} (x^{3} + a x y^{2})

x \to - \infty lim (\frac{4 arctan ( 8 x )}{e ^{x} + 2})

\int x d x

\frac{\partial}{\partial y} (6 y^{5} cos (5 x))

What is the integral of (2-v)/(v^2+2v-3) ?
The integral of (2-v)/(v^2+2v-3) is 1/4 ln|v-1|-5/4 ln|v+3|+C