What is the integral of (x+2)/(x^2+4x-1) ?

The integral of (x+2)/(x^2+4x-1) is

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integral of (x+2)/(x^2+4x-1)

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

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

Solution steps

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

Expand

\frac{x + 2}{x ^{2} + 4 x - 1} : \frac{x}{x ^{2} + 4 x - 1} + \frac{2}{x ^{2} + 4 x - 1}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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