What is the integral of (-x+7)/(x^2+x-6) ?

The integral of (-x+7)/(x^2+x-6) is

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integral of (-x+7)/(x^2+x-6)

\int \frac{- x + 7}{x ^{2} + x - 6} d x

- 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{7}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) + C

Solution steps

\int \frac{- x + 7}{x ^{2} + x - 6} d x

Expand

\frac{- x + 7}{x ^{2} + x - 6} : - \frac{x}{x ^{2} + x - 6} + \frac{7}{x ^{2} + x - 6}

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{x}{x ^{2} + x - 6} d x + \int \frac{7}{x ^{2} + x - 6} d x

\int \frac{x}{x ^{2} + x - 6} d x = 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

\int \frac{7}{x ^{2} + x - 6} d x = - \frac{7}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5})

= - 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{7}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5})

Add a constant to the solution

= - 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{7}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) + C

\int \frac{x ^{2}}{x ^{2} - 4 x + 3} d x

x^{\frac{3}{2}}, 4

\frac{\partial}{\partial x} (- 210 (5 x - y)^{5})

\frac{d}{d x} (1 + c 1 - x^{2})

y = \frac{6 x}{x ^{2} + 1}