What is the integral of (3x-2)/(x^2-x-3) ?

The integral of (3x-2)/(x^2-x-3) is

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integral of (3x-2)/(x^2-x-3)

\int \frac{3 x - 2}{x ^{2} - x - 3} d x

\frac{3}{2} ln 4 x^{2} - 4 x - 12 + \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) - 1 + C

Solution steps

\int \frac{3 x - 2}{x ^{2} - x - 3} d x

Expand

\frac{3 x - 2}{x ^{2} - x - 3} : \frac{3 x}{x ^{2} - x - 3} - \frac{2}{x ^{2} - x - 3}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 x}{x ^{2} - x - 3} d x - \int \frac{2}{x ^{2} - x - 3} d x

\int \frac{3 x}{x ^{2} - x - 3} d x = 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 12 - \frac{1}{4 13} (ln \frac{2 x - 1}{13} + 1 - ln \frac{2 x - 1}{13} - 1))

\int \frac{2}{x ^{2} - x - 3} d x = - \frac{2}{13} (ln \frac{2 x - 1}{13} + 1 - ln \frac{2 x - 1}{13} - 1)

= 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 12 - \frac{1}{4 13} (ln \frac{2 x - 1}{13} + 1 - ln \frac{2 x - 1}{13} - 1)) - (- \frac{2}{13} (ln \frac{2 x - 1}{13} + 1 - ln \frac{2 x - 1}{13} - 1))

Simplify

6 (\frac{1}{4} ln 4 x^{2} - 4 x - 12 - \frac{1}{4 13} (ln \frac{2 x - 1}{13} + 1 - ln \frac{2 x - 1}{13} - 1)) - (- \frac{2}{13} (ln \frac{2 x - 1}{13} + 1 - ln \frac{2 x - 1}{13} - 1)) : \frac{3}{2} ln 4 x^{2} - 4 x - 12 + \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) - 1

= \frac{3}{2} ln 4 x^{2} - 4 x - 12 + \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) - 1

Add a constant to the solution

= \frac{3}{2} ln 4 x^{2} - 4 x - 12 + \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 x - 1) - 1 + C

\int 14 x - x^{2} d x

\frac{d}{d x} (x^{\frac{- 2}{5}})

f (x) = 2 x cos (x^{2})

\int_{0}^{2} \frac{7}{x - 2} d x

k = 1 \sum \infty ln (k)