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

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

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

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

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

Solution steps

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

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 x + 1

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

Simplify

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

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

Add a constant to the solution

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

2 x e^{- x}

\frac{d}{d x} (\frac{5}{12} x^{\frac{6}{5}} - \frac{5}{8} x^{\frac{4}{5}})

\frac{\partial}{\partial y} (5 x^{2} - 2 y)

\frac{1}{( s ^{2} + 9 ) ^{2}}

n = 1 \sum \infty \frac{9}{n ln ( 2 n )}

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