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

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

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

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

- ln ∣ x - 1 ∣ + 3 ln ∣ x - 3 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

= - ln ∣ x - 1 ∣ + 3 ln ∣ x - 3 ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + 3 ln ∣ x - 3 ∣ + C

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