What is the integral of ((x-1))/(x^3+x^2-4x-4) ?

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

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integral of ((x-1))/(x^3+x^2-4x-4)

\int \frac{( x - 1 )}{x ^{3} + x ^{2} - 4 x - 4} d x

\frac{2}{3} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x - 1}{x ^{3} + x ^{2} - 4 x - 4} d x

Take the partial fraction of

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

= \int \frac{2}{3 ( x + 1 )} - \frac{3}{4 ( x + 2 )} + \frac{1}{12 ( 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 \frac{2}{3 ( x + 1 )} d x - \int \frac{3}{4 ( x + 2 )} d x + \int \frac{1}{12 ( x - 2 )} d x

\int \frac{2}{3 ( x + 1 )} d x = \frac{2}{3} ln ∣ x + 1 ∣

\int \frac{3}{4 ( x + 2 )} d x = \frac{3}{4} ln ∣ x + 2 ∣

\int \frac{1}{12 ( x - 2 )} d x = \frac{1}{12} ln ∣ x - 2 ∣

= \frac{2}{3} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{2}{3} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣ + C

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