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

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

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integral of 1/(y^2-y-2)

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

- \frac{1}{3} (ln \frac{2}{3} + \frac{2}{3} y - ln - \frac{4}{3} + \frac{2}{3} y) + C

Solution steps

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

Complete the square

y^{2} - y - 2 : (y - \frac{1}{2})^{2} - \frac{9}{4}

= \int \frac{1}{( y - \frac{1}{2} ) ^{2} - \frac{9}{4}} d y

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 9} d u

= \int \frac{4}{4 u ^{2} - 9} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} - 9} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{6 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{6} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

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

= 4 \cdot \frac{1}{6} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( y - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( y - \frac{1}{2} ) - 1}{2}))

Simplify

4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( y - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( y - \frac{1}{2} ) - 1}{2})) : - \frac{1}{3} (ln \frac{2}{3} + \frac{2}{3} y - ln - \frac{4}{3} + \frac{2}{3} y)

= - \frac{1}{3} (ln \frac{2}{3} + \frac{2}{3} y - ln - \frac{4}{3} + \frac{2}{3} y)

Add a constant to the solution

= - \frac{1}{3} (ln \frac{2}{3} + \frac{2}{3} y - ln - \frac{4}{3} + \frac{2}{3} y) + C

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