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

The integral of (x^2+x-2)/((x+2)(x-5)(x+1)) is 2/3 ln|x-5|+1/3 ln|x+1|+C

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

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

\frac{2}{3} ln ∣ x - 5 ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + C

Solution steps

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

Simplify

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

= \int \frac{x - 1}{( x - 5 ) ( x + 1 )} d x

Take the partial fraction of

\frac{x - 1}{( x - 5 ) ( x + 1 )} : \frac{2}{3 ( x - 5 )} + \frac{1}{3 ( x + 1 )}

= \int \frac{2}{3 ( x - 5 )} + \frac{1}{3 ( x + 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{2}{3 ( x - 5 )} d x + \int \frac{1}{3 ( x + 1 )} d x

\int \frac{2}{3 ( x - 5 )} d x = \frac{2}{3} ln ∣ x - 5 ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

= \frac{2}{3} ln ∣ x - 5 ∣ + \frac{1}{3} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{2}{3} ln ∣ x - 5 ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + C

\frac{\partial}{\partial x} (2 sin (x y))

f (x) = - x^{3} + 15 x^{2} - 48 x + 450

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

\frac{\partial}{\partial x} (\frac{x ^{2}}{x ^{2} - 1})

f (x) = 4 x, at x = 1

What is the integral of (x^2+x-2)/((x+2)(x-5)(x+1)) ?
The integral of (x^2+x-2)/((x+2)(x-5)(x+1)) is 2/3 ln|x-5|+1/3 ln|x+1|+C