What is the integral of (24-x)/(x^2+2x-24) ?

The integral of (24-x)/(x^2+2x-24) is

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integral of (24-x)/(x^2+2x-24)

\int \frac{24 - x}{x ^{2} + 2 x - 24} d x

- \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1) + C

Solution steps

\int \frac{24 - x}{x ^{2} + 2 x - 24} d x

Expand

\frac{24 - x}{x ^{2} + 2 x - 24} : \frac{24}{x ^{2} + 2 x - 24} - \frac{x}{x ^{2} + 2 x - 24}

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{24}{x ^{2} + 2 x - 24} d x - \int \frac{x}{x ^{2} + 2 x - 24} d x

\int \frac{24}{x ^{2} + 2 x - 24} d x = - \frac{12}{5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1)

\int \frac{x}{x ^{2} + 2 x - 24} d x = \frac{1}{2} ln x^{2} + 2 x - 24 + \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)

= - \frac{12}{5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1) - (\frac{1}{2} ln x^{2} + 2 x - 24 + \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1))

Simplify

- \frac{12}{5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1) - (\frac{1}{2} ln x^{2} + 2 x - 24 + \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)) : - \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)

= - \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)

Add a constant to the solution

= - \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1) + C

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What is the integral of (24-x)/(x^2+2x-24) ?
The integral of (24-x)/(x^2+2x-24) is