What is the integral of ((2x+4))/(x^2+5x-6) ?

The integral of ((2x+4))/(x^2+5x-6) is 6/7 ln|x-1|+8/7 ln|x+6|+C

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integral of ((2x+4))/(x^2+5x-6)

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

\frac{6}{7} ln ∣ x - 1 ∣ + \frac{8}{7} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{2 x + 4}{x ^{2} + 5 x - 6} d x

Expand

\frac{2 x + 4}{x ^{2} + 5 x - 6} : \frac{2 x}{x ^{2} + 5 x - 6} + \frac{4}{x ^{2} + 5 x - 6}

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} + 5 x - 6} d x + \int \frac{4}{x ^{2} + 5 x - 6} d x

\int \frac{2 x}{x ^{2} + 5 x - 6} d x = 2 (\frac{1}{7} ln ∣ x - 1 ∣ + \frac{6}{7} ln ∣ x + 6 ∣)

\int \frac{4}{x ^{2} + 5 x - 6} d x = 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣)

= 2 (\frac{1}{7} ln ∣ x - 1 ∣ + \frac{6}{7} ln ∣ x + 6 ∣) + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣)

Simplify

2 (\frac{1}{7} ln ∣ x - 1 ∣ + \frac{6}{7} ln ∣ x + 6 ∣) + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣) : \frac{6}{7} ln ∣ x - 1 ∣ + \frac{8}{7} ln ∣ x + 6 ∣

= \frac{6}{7} ln ∣ x - 1 ∣ + \frac{8}{7} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{6}{7} ln ∣ x - 1 ∣ + \frac{8}{7} ln ∣ x + 6 ∣ + C

\int \frac{1}{t ^{2} - 6 t + 25} d t

\frac{d}{d x} (cos (ln (3 x)))

\int x - x d x

2 - x^{2}, - x

x \to - 1 lim (4 x^{2})

What is the integral of ((2x+4))/(x^2+5x-6) ?
The integral of ((2x+4))/(x^2+5x-6) is 6/7 ln|x-1|+8/7 ln|x+6|+C