What is the integral of (2x+1)/(x^2+2x-15) ?

The integral of (2x+1)/(x^2+2x-15) is

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integral of (2x+1)/(x^2+2x-15)

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

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

Solution steps

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

Expand

\frac{2 x + 1}{x ^{2} + 2 x - 15} : \frac{2 x}{x ^{2} + 2 x - 15} + \frac{1}{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{2 x}{x ^{2} + 2 x - 15} d x + \int \frac{1}{x ^{2} + 2 x - 15} d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (2x+1)/(x^2+2x-15) ?
The integral of (2x+1)/(x^2+2x-15) is