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

The integral of ((7x-1))/((x-1)(x-2)(x-3)) is 3ln|x-1|+10ln|x-3|-13ln|x-2|+C

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

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

3 ln ∣ x - 1 ∣ + 10 ln ∣ x - 3 ∣ - 13 ln ∣ x - 2 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{7 x}{( x - 1 ) ( x - 2 ) ( x - 3 )} d x = 7 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ - 2 ln ∣ x - 2 ∣)

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

= 7 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ - 2 ln ∣ x - 2 ∣) - (\frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣)

Simplify

7 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ - 2 ln ∣ x - 2 ∣) - (\frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣) : 3 ln ∣ x - 1 ∣ + 10 ln ∣ x - 3 ∣ - 13 ln ∣ x - 2 ∣

= 3 ln ∣ x - 1 ∣ + 10 ln ∣ x - 3 ∣ - 13 ln ∣ x - 2 ∣

Add a constant to the solution

= 3 ln ∣ x - 1 ∣ + 10 ln ∣ x - 3 ∣ - 13 ln ∣ x - 2 ∣ + C

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What is the integral of ((7x-1))/((x-1)(x-2)(x-3)) ?
The integral of ((7x-1))/((x-1)(x-2)(x-3)) is 3ln|x-1|+10ln|x-3|-13ln|x-2|+C