What is the integral of (2x-1)/(x^2+4x-5) ?

The integral of (2x-1)/(x^2+4x-5) is 1/6 ln|x-1|+11/6 ln|x+5|+C

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integral of (2x-1)/(x^2+4x-5)

\int \frac{2 x - 1}{x ^{2} + 4 x - 5} d x

\frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{2 x - 1}{x ^{2} + 4 x - 5} d x

Expand

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

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} + 4 x - 5} d x - \int \frac{1}{x ^{2} + 4 x - 5} d x

\int \frac{2 x}{x ^{2} + 4 x - 5} d x = 2 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣)

\int \frac{1}{x ^{2} + 4 x - 5} d x = \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣

= 2 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣) - (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣)

Simplify

2 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣) - (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) : \frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x + 5 ∣

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x + 5 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x + 5 ∣ + C

What is the integral of (2x-1)/(x^2+4x-5) ?
The integral of (2x-1)/(x^2+4x-5) is 1/6 ln|x-1|+11/6 ln|x+5|+C