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

The integral of (3x-2)/(x^2+x-2) is 1/3 ln|x-1|+8/3 ln|x+2|+C

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

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

\frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{3 x - 2}{x ^{2} + x - 2} : \frac{1}{3 ( x - 1 )} + \frac{8}{3 ( x + 2 )}

= \int \frac{1}{3 ( x - 1 )} + \frac{8}{3 ( 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{1}{3 ( x - 1 )} d x + \int \frac{8}{3 ( x + 2 )} d x

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

\int \frac{8}{3 ( x + 2 )} d x = \frac{8}{3} ln ∣ x + 2 ∣

= \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ + C

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What is the integral of (3x-2)/(x^2+x-2) ?
The integral of (3x-2)/(x^2+x-2) is 1/3 ln|x-1|+8/3 ln|x+2|+C