What is the integral of ((5y+1))/((2y^2-y-1)) ?

The integral of ((5y+1))/((2y^2-y-1)) is

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integral of ((5y+1))/((2y^2-y-1))

\int \frac{( 5 y + 1 )}{( 2 y ^{2} - y - 1 )} d y

- \frac{1}{3} (5 y + 1) (ln \frac{4}{15} (5 y + 1) + \frac{2}{5} - ln \frac{4}{15} (5 y + 1) - \frac{8}{5}) + \frac{1}{3} ln \frac{4}{15} (5 y + 1) + \frac{2}{5} (5 y + 1) - \frac{1}{3} ln \frac{4}{15} (5 y + 1) - \frac{8}{5} (5 y + 1) + \frac{1}{2} ln ∣ 10 y + 5 ∣ + 2 ln ∣ 5 y - 5 ∣ + C

Solution steps

\int \frac{5 y + 1}{2 y ^{2} - y - 1} d y

Apply u-substitution

: \int \frac{5 u}{2 u ^{2} - 9 u - 18} d u

= \int \frac{5 u}{2 u ^{2} - 9 u - 18} d u

Take the constant out:

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

= 5 \cdot \int \frac{u}{2 u ^{2} - 9 u - 18} d u

Apply Integration By Parts

: - \frac{1}{15} u (ln \frac{2}{5} + \frac{4}{15} u - ln - \frac{8}{5} + \frac{4}{15} u) - \int - \frac{1}{15} (ln \frac{2}{5} + \frac{4}{15} u - ln - \frac{8}{5} + \frac{4}{15} u) d u

= 5 (- \frac{1}{15} u (ln \frac{2}{5} + \frac{4}{15} u - ln - \frac{8}{5} + \frac{4}{15} u) - \int - \frac{1}{15} (ln \frac{2}{5} + \frac{4}{15} u - ln - \frac{8}{5} + \frac{4}{15} u) d u)

\int - \frac{1}{15} (ln \frac{2}{5} + \frac{4}{15} u - ln - \frac{8}{5} + \frac{4}{15} u) d u = - \frac{1}{15} u ln \frac{2}{5} + \frac{4}{15} u + \frac{1}{15} u ln - \frac{8}{5} + \frac{4}{15} u - \frac{1}{10} ln ∣ 2 u + 3 ∣ - \frac{2}{5} ln ∣ u - 6 ∣

= 5 (- \frac{1}{15} u (ln \frac{2}{5} + \frac{4}{15} u - ln - \frac{8}{5} + \frac{4}{15} u) - (- \frac{1}{15} u ln \frac{2}{5} + \frac{4}{15} u + \frac{1}{15} u ln - \frac{8}{5} + \frac{4}{15} u - \frac{1}{10} ln ∣ 2 u + 3 ∣ - \frac{2}{5} ln ∣ u - 6 ∣))

Substitute back

u = 5 y + 1

= 5 (- \frac{1}{15} (5 y + 1) (ln \frac{2}{5} + \frac{4}{15} (5 y + 1) - ln - \frac{8}{5} + \frac{4}{15} (5 y + 1)) - (- \frac{1}{15} (5 y + 1) ln \frac{2}{5} + \frac{4}{15} (5 y + 1) + \frac{1}{15} (5 y + 1) ln - \frac{8}{5} + \frac{4}{15} (5 y + 1) - \frac{1}{10} ln ∣ 2 (5 y + 1) + 3 ∣ - \frac{2}{5} ln ∣ 5 y + 1 - 6 ∣))

Simplify

5 (- \frac{1}{15} (5 y + 1) (ln \frac{2}{5} + \frac{4}{15} (5 y + 1) - ln - \frac{8}{5} + \frac{4}{15} (5 y + 1)) - (- \frac{1}{15} (5 y + 1) ln \frac{2}{5} + \frac{4}{15} (5 y + 1) + \frac{1}{15} (5 y + 1) ln - \frac{8}{5} + \frac{4}{15} (5 y + 1) - \frac{1}{10} ln ∣ 2 (5 y + 1) + 3 ∣ - \frac{2}{5} ln ∣ 5 y + 1 - 6 ∣)) : - \frac{1}{3} (5 y + 1) (ln \frac{4}{15} (5 y + 1) + \frac{2}{5} - ln \frac{4}{15} (5 y + 1) - \frac{8}{5}) + \frac{1}{3} ln \frac{4}{15} (5 y + 1) + \frac{2}{5} (5 y + 1) - \frac{1}{3} ln \frac{4}{15} (5 y + 1) - \frac{8}{5} (5 y + 1) + \frac{1}{2} ln ∣ 10 y + 5 ∣ + 2 ln ∣ 5 y - 5 ∣

= - \frac{1}{3} (5 y + 1) (ln \frac{4}{15} (5 y + 1) + \frac{2}{5} - ln \frac{4}{15} (5 y + 1) - \frac{8}{5}) + \frac{1}{3} ln \frac{4}{15} (5 y + 1) + \frac{2}{5} (5 y + 1) - \frac{1}{3} ln \frac{4}{15} (5 y + 1) - \frac{8}{5} (5 y + 1) + \frac{1}{2} ln ∣ 10 y + 5 ∣ + 2 ln ∣ 5 y - 5 ∣

Add a constant to the solution

= - \frac{1}{3} (5 y + 1) (ln \frac{4}{15} (5 y + 1) + \frac{2}{5} - ln \frac{4}{15} (5 y + 1) - \frac{8}{5}) + \frac{1}{3} ln \frac{4}{15} (5 y + 1) + \frac{2}{5} (5 y + 1) - \frac{1}{3} ln \frac{4}{15} (5 y + 1) - \frac{8}{5} (5 y + 1) + \frac{1}{2} ln ∣ 10 y + 5 ∣ + 2 ln ∣ 5 y - 5 ∣ + C

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What is the integral of ((5y+1))/((2y^2-y-1)) ?
The integral of ((5y+1))/((2y^2-y-1)) is