What is the integral of (v+3)/(v-1) ?

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

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integral of (v+3)/(v-1)

\int \frac{v + 3}{v - 1} d v

v - 1 + 4 ln ∣ v - 1 ∣ + C

Solution steps

\int \frac{v + 3}{v - 1} d v

Expand

\frac{v + 3}{v - 1} : \frac{v}{v - 1} + \frac{3}{v - 1}

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{v}{v - 1} d v + \int \frac{3}{v - 1} d v

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

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

= v - 1 + ln ∣ v - 1 ∣ + 3 ln ∣ v - 1 ∣

Simplify

= v - 1 + 4 ln ∣ v - 1 ∣

Add a constant to the solution

= v - 1 + 4 ln ∣ v - 1 ∣ + C

\frac{d}{d x} (3 (5 - x)^{2})

\frac{\partial}{\partial z} (z^{2} cos (y x))

f (x) = (4 t - 1) (3 t - 6)^{- 1}

\int \frac{x ^{2}}{( x ^{3} - 6 ) ^{2}} d x

f (x) = \frac{1}{x ^{10}}

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