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

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

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integral of 1/(1-v^2)

\int \frac{1}{1 - v ^{2}} d v

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

Solution steps

\int \frac{1}{1 - v ^{2}} d v

Use the common integral:

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

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

Add a constant to the solution

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

\frac{d}{d t} (1 - ∣ t ∣)

\int - sin (x) - e^{- x} + 3 x^{2} d x

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + e^{- x}

\frac{d y}{d x} = 2 x (y - 1)

\int \frac{7 - x ^{4}}{5 x} d x

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