What is the integral of 1/(v^2+2v-3) ?

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

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integral of 1/(v^2+2v-3)

\int \frac{1}{v ^{2} + 2 v - 3} d v

- \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1) + C

Solution steps

\int \frac{1}{v ^{2} + 2 v - 3} d v

Factor

v^{2} + 2 v - 3 : - (- v^{2} - 2 v + 3)

= \int \frac{1}{- ( - v ^{2} - 2 v + 3 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

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

= - \int \frac{1}{- ( v + 1 ) ^{2} + 4} d v

Apply u-substitution

= - \int \frac{1}{- u ^{2} + 4} d u

Apply Integral Substitution

= - \int \frac{1}{2 ( - w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{- w ^{2} + 1} d w

Use the common integral:

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

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

Substitute back

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

Simplify

- \frac{1}{2} (\frac{ln \frac{v + 1}{2} + 1}{2} - \frac{ln \frac{v + 1}{2} - 1}{2}) : - \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1)

= - \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1)

Add a constant to the solution

= - \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1) + C

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