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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{5}{x + 2} d x = 5 ln ∣ x + 2 ∣

\int \frac{5}{2 ( x + 1 )} d x = \frac{5}{2} ln ∣ x + 1 ∣

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

Add a constant to the solution

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

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