What is the integral of (x-4)/(x^2+2x-15) ?

The integral of (x-4)/(x^2+2x-15) is

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integral of (x-4)/(x^2+2x-15)

\int \frac{x - 4}{x ^{2} + 2 x - 15} d x

\frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1) + \frac{1}{2} ln \frac{1}{4} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{4} (x + 1) - 1 + C

Solution steps

\int \frac{x - 4}{x ^{2} + 2 x - 15} d x

Expand

\frac{x - 4}{x ^{2} + 2 x - 15} : \frac{x}{x ^{2} + 2 x - 15} - \frac{4}{x ^{2} + 2 x - 15}

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} + 2 x - 15} d x - \int \frac{4}{x ^{2} + 2 x - 15} d x

\int \frac{x}{x ^{2} + 2 x - 15} d x = \frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1)

\int \frac{4}{x ^{2} + 2 x - 15} d x = - \frac{1}{2} ln \frac{x + 1}{4} + 1 + \frac{1}{2} ln \frac{x + 1}{4} - 1

= \frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1) - (- \frac{1}{2} ln \frac{x + 1}{4} + 1 + \frac{1}{2} ln \frac{x + 1}{4} - 1)

Simplify

\frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1) - (- \frac{1}{2} ln \frac{x + 1}{4} + 1 + \frac{1}{2} ln \frac{x + 1}{4} - 1) : \frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1) + \frac{1}{2} ln \frac{1}{4} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{4} (x + 1) - 1

= \frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1) + \frac{1}{2} ln \frac{1}{4} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{4} (x + 1) - 1

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x - 15 + \frac{1}{8} (ln \frac{1}{4} (x + 1) + 1 - ln \frac{1}{4} (x + 1) - 1) + \frac{1}{2} ln \frac{1}{4} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{4} (x + 1) - 1 + C

What is the integral of (x-4)/(x^2+2x-15) ?
The integral of (x-4)/(x^2+2x-15) is