What is the integral of (5x^2-6)/(x^2-2x-8) ?

The integral of (5x^2-6)/(x^2-2x-8) is 5x-7/3 ln|x+2|+37/3 ln|x-4|+C

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integral of (5x^2-6)/(x^2-2x-8)

\int \frac{5 x ^{2} - 6}{x ^{2} - 2 x - 8} d x

5 x - \frac{7}{3} ln ∣ x + 2 ∣ + \frac{37}{3} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{5 x ^{2} - 6}{x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{5 x ^{2} - 6}{x ^{2} - 2 x - 8} : 5 - \frac{7}{3 ( x + 2 )} + \frac{37}{3 ( x - 4 )}

= \int 5 - \frac{7}{3 ( x + 2 )} + \frac{37}{3 ( x - 4 )} 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 5 d x - \int \frac{7}{3 ( x + 2 )} d x + \int \frac{37}{3 ( x - 4 )} d x

\int 5 d x = 5 x

\int \frac{7}{3 ( x + 2 )} d x = \frac{7}{3} ln ∣ x + 2 ∣

\int \frac{37}{3 ( x - 4 )} d x = \frac{37}{3} ln ∣ x - 4 ∣

= 5 x - \frac{7}{3} ln ∣ x + 2 ∣ + \frac{37}{3} ln ∣ x - 4 ∣

Add a constant to the solution

= 5 x - \frac{7}{3} ln ∣ x + 2 ∣ + \frac{37}{3} ln ∣ x - 4 ∣ + C

\frac{d}{d x} (sin^{2} (3 x + 2))

f (x) = c y^{- 7}

f (x) = x^{2} + x, at x = 1, x = 5

(x^{3})^{^{'}}

\frac{d y}{d x} = \frac{( y ^{2} )}{x}

What is the integral of (5x^2-6)/(x^2-2x-8) ?
The integral of (5x^2-6)/(x^2-2x-8) is 5x-7/3 ln|x+2|+37/3 ln|x-4|+C