What is the integral of (x^2+3x-4)/(x^2-2x-8) ?

The integral of (x^2+3x-4)/(x^2-2x-8) is ln|x+2|+4ln|x-4|+x+C

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integral of (x^2+3x-4)/(x^2-2x-8)

\int \frac{x ^{2} + 3 x - 4}{x ^{2} - 2 x - 8} d x

ln ∣ x + 2 ∣ + 4 ln ∣ x - 4 ∣ + x + C

Solution steps

\int \frac{x ^{2} + 3 x - 4}{x ^{2} - 2 x - 8} d x

\frac{x ^{2} + 3 x - 4}{x ^{2} - 2 x - 8} = \frac{5 x + 4}{x ^{2} - 2 x - 8} + 1

= \int \frac{5 x + 4}{x ^{2} - 2 x - 8} + 1 d x

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{5 x + 4}{x ^{2} - 2 x - 8} d x + \int 1 d x

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

\int 1 d x = x

= ln ∣ x + 2 ∣ + 4 ln ∣ x - 4 ∣ + x

Add a constant to the solution

= ln ∣ x + 2 ∣ + 4 ln ∣ x - 4 ∣ + x + C

\int (y^{2} - 1) d y

x^{3} - 13 x^{2} + 40 x, - x^{3} + 13 x^{2} - 40 x

\int 900 - 36 x^{2} d x

y^{^{'}} = 13 x y - 16 x

\int \frac{( 2 x ^{3} + x - 1 )}{x ^{3} + x ^{2} - 4 x - 4} d x

What is the integral of (x^2+3x-4)/(x^2-2x-8) ?
The integral of (x^2+3x-4)/(x^2-2x-8) is ln|x+2|+4ln|x-4|+x+C