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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{4 ( x - 3 )} - \frac{1}{4 ( 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

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

\int \frac{1}{4 ( x - 3 )} d x = \frac{1}{4} ln ∣ x - 3 ∣

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

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

Add a constant to the solution

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

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