What is the integral of ((x^2+4x-8))/(x^2+4x-5) ?

The integral of ((x^2+4x-8))/(x^2+4x-5) is x-1/2 ln|x-1|+1/2 ln|x+5|+C

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integral of ((x^2+4x-8))/(x^2+4x-5)

\int \frac{( x ^{2} + 4 x - 8 )}{x ^{2} + 4 x - 5} d x

x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{x ^{2} + 4 x - 8}{x ^{2} + 4 x - 5} d x

Expand

\frac{x ^{2} + 4 x - 8}{x ^{2} + 4 x - 5} : \frac{x ^{2}}{x ^{2} + 4 x - 5} + \frac{4 x}{x ^{2} + 4 x - 5} - \frac{8}{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{x ^{2}}{x ^{2} + 4 x - 5} d x + \int \frac{4 x}{x ^{2} + 4 x - 5} d x - \int \frac{8}{x ^{2} + 4 x - 5} d x

\int \frac{x ^{2}}{x ^{2} + 4 x - 5} d x = x + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{25}{6} ln ∣ x + 5 ∣

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

\int \frac{8}{x ^{2} + 4 x - 5} d x = 8 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣)

= x + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{25}{6} ln ∣ x + 5 ∣ + 4 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣) - 8 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣)

Simplify

x + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{25}{6} ln ∣ x + 5 ∣ + 4 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣) - 8 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) : x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x + 5 ∣

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

Add a constant to the solution

= x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x + 5 ∣ + C

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What is the integral of ((x^2+4x-8))/(x^2+4x-5) ?
The integral of ((x^2+4x-8))/(x^2+4x-5) is x-1/2 ln|x-1|+1/2 ln|x+5|+C