What is the integral of (x^3-4x-10)/(x^3-3x^2-4x+12) ?

The integral of (x^3-4x-10)/(x^3-3x^2-4x+12) is x+ln|x-3|-1/2 ln|x+2|+5/2 ln|x-2|+C

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integral of (x^3-4x-10)/(x^3-3x^2-4x+12)

\int \frac{x ^{3} - 4 x - 10}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

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

Solution steps

\int \frac{x ^{3} - 4 x - 10}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

Take the partial fraction of

\frac{x ^{3} - 4 x - 10}{x ^{3} - 3 x ^{2} - 4 x + 12} : 1 + \frac{1}{x - 3} - \frac{1}{2 ( x + 2 )} + \frac{5}{2 ( x - 2 )}

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

\int 1 d x = x

\int \frac{1}{x - 3} d x = ln ∣ x - 3 ∣

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

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

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

Add a constant to the solution

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

(x^{2} - 1)^{^{'}}

\int \frac{x}{2 x ^{2} - 1} d x

\frac{d y}{d x} - (\frac{4}{x}) y = x^{5} e^{x}

x \to 2 lim (\frac{x ^{2} - 2}{\frac{5 x}{7} - \frac{5 2}{7}})

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

What is the integral of (x^3-4x-10)/(x^3-3x^2-4x+12) ?
The integral of (x^3-4x-10)/(x^3-3x^2-4x+12) is x+ln|x-3|-1/2 ln|x+2|+5/2 ln|x-2|+C