What is the integral of (x^2-4x+5)/((x^2-1)(x+3)) ?

The integral of (x^2-4x+5)/((x^2-1)(x+3)) is -5/2 ln|x+1|+1/4 ln|x-1|+13/4 ln|x+3|+C

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integral of (x^2-4x+5)/((x^2-1)(x+3))

\int \frac{x ^{2} - 4 x + 5}{( x ^{2} - 1 ) ( x + 3 )} d x

- \frac{5}{2} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{13}{4} ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{x ^{2} - 4 x + 5}{( x ^{2} - 1 ) ( x + 3 )} d x

Take the partial fraction of

\frac{x ^{2} - 4 x + 5}{( x ^{2} - 1 ) ( x + 3 )} : - \frac{5}{2 ( x + 1 )} + \frac{1}{4 ( x - 1 )} + \frac{13}{4 ( x + 3 )}

= \int - \frac{5}{2 ( x + 1 )} + \frac{1}{4 ( x - 1 )} + \frac{13}{4 ( 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{5}{2 ( x + 1 )} d x + \int \frac{1}{4 ( x - 1 )} d x + \int \frac{13}{4 ( x + 3 )} d x

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{13}{4 ( x + 3 )} d x = \frac{13}{4} ln ∣ x + 3 ∣

= - \frac{5}{2} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{13}{4} ln ∣ x + 3 ∣

Add a constant to the solution

= - \frac{5}{2} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{13}{4} ln ∣ x + 3 ∣ + C

What is the integral of (x^2-4x+5)/((x^2-1)(x+3)) ?
The integral of (x^2-4x+5)/((x^2-1)(x+3)) is -5/2 ln|x+1|+1/4 ln|x-1|+13/4 ln|x+3|+C