What is the integral of (x+7)/(x^2-8x-19) ?

The integral of (x+7)/(x^2-8x-19) is

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integral of (x+7)/(x^2-8x-19)

\int \frac{x + 7}{x ^{2} - 8 x - 19} d x

\frac{1}{2} ln x^{2} - 8 x - 19 - \frac{2}{35} (ln \frac{1}{35} (x - 4) + 1 - ln \frac{1}{35} (x - 4) - 1) - \frac{7}{2 5} (ln \frac{x - 4}{35} + 1 - ln \frac{x - 4}{35} - 1) + C

Solution steps

\int \frac{x + 7}{x ^{2} - 8 x - 19} d x

Expand

\frac{x + 7}{x ^{2} - 8 x - 19} : \frac{x}{x ^{2} - 8 x - 19} + \frac{7}{x ^{2} - 8 x - 19}

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} - 8 x - 19} d x + \int \frac{7}{x ^{2} - 8 x - 19} d x

\int \frac{x}{x ^{2} - 8 x - 19} d x = \frac{1}{2} ln x^{2} - 8 x - 19 - \frac{2}{35} (ln \frac{1}{35} (x - 4) + 1 - ln \frac{1}{35} (x - 4) - 1)

\int \frac{7}{x ^{2} - 8 x - 19} d x = - \frac{7}{2 5} (ln \frac{x - 4}{35} + 1 - ln \frac{x - 4}{35} - 1)

= \frac{1}{2} ln x^{2} - 8 x - 19 - \frac{2}{35} (ln \frac{1}{35} (x - 4) + 1 - ln \frac{1}{35} (x - 4) - 1) - \frac{7}{2 5} (ln \frac{x - 4}{35} + 1 - ln \frac{x - 4}{35} - 1)

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 8 x - 19 - \frac{2}{35} (ln \frac{1}{35} (x - 4) + 1 - ln \frac{1}{35} (x - 4) - 1) - \frac{7}{2 5} (ln \frac{x - 4}{35} + 1 - ln \frac{x - 4}{35} - 1) + C

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What is the integral of (x+7)/(x^2-8x-19) ?
The integral of (x+7)/(x^2-8x-19) is