What is the integral of (x+5)/(x^3+2x^2-x-2) ?

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

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integral of (x+5)/(x^3+2x^2-x-2)

\int \frac{x + 5}{x ^{3} + 2 x ^{2} - x - 2} d x

ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x + 5}{x ^{3} + 2 x ^{2} - x - 2} d x

Take the partial fraction of

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

= \int \frac{1}{x + 2} - \frac{2}{x + 1} + \frac{1}{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{1}{x + 2} d x - \int \frac{2}{x + 1} d x + \int \frac{1}{x - 1} d x

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

2 y^{^{'}} = y

\frac{\partial}{\partial x} (4 sin^{2} (x))

f (x) = \frac{2}{x}

e^{π}

\frac{\partial}{\partial y} (4 x - 2 y)

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