What is the integral of (5x-7)/((x^2-1)(x-3)) ?

The integral of (5x-7)/((x^2-1)(x-3)) is -3/2 ln|x+1|+1/2 ln|x-1|+ln|x-3|+C

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integral of (5x-7)/((x^2-1)(x-3))

\int \frac{5 x - 7}{( x ^{2} - 1 ) ( x - 3 )} d x

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

Solution steps

\int \frac{5 x - 7}{( x ^{2} - 1 ) ( x - 3 )} d x

Take the partial fraction of

\frac{5 x - 7}{( x ^{2} - 1 ) ( x - 3 )} : - \frac{3}{2 ( x + 1 )} + \frac{1}{2 ( x - 1 )} + \frac{1}{x - 3}

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

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

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

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

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

Add a constant to the solution

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

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What is the integral of (5x-7)/((x^2-1)(x-3)) ?
The integral of (5x-7)/((x^2-1)(x-3)) is -3/2 ln|x+1|+1/2 ln|x-1|+ln|x-3|+C