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Popular >

integral of (x+1)/(x^2+x-6)

Solution

\int \frac{x + 1}{x ^{2} + x - 6} d x

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

Solution steps

\int \frac{x + 1}{x ^{2} + x - 6} d x

Expand

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

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

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Frequently Asked Questions (FAQ)

What is the integral of (x+1)/(x^2+x-6) ?
The integral of (x+1)/(x^2+x-6) is