What is the integral of ((x^3-4x-10))/((x^2-x-6)) ?

The integral of ((x^3-4x-10))/((x^2-x-6)) is (x^2)/2+x+2ln|x+2|+ln|x-3|+C

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integral of ((x^3-4x-10))/((x^2-x-6))

\int \frac{( x ^{3} - 4 x - 10 )}{( x ^{2} - x - 6 )} d x

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

Solution steps

\int \frac{x ^{3} - 4 x - 10}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{x ^{3} - 4 x - 10}{x ^{2} - x - 6} : x + 1 + \frac{2}{x + 2} + \frac{1}{x - 3}

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

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

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

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

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

Add a constant to the solution

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

\int \frac{5}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{d}{d x} (- \frac{7}{x ^{4}})

y = \frac{x}{x - 1}

\int_{0}^{1} 1 - x^{\frac{2}{5}} d x

f (x) = x^{3} - x^{2} - x, g (x) = x

What is the integral of ((x^3-4x-10))/((x^2-x-6)) ?
The integral of ((x^3-4x-10))/((x^2-x-6)) is (x^2)/2+x+2ln|x+2|+ln|x-3|+C