What is the integral of x/((x-1)(x-4)(x-1)) ?

The integral of x/((x-1)(x-4)(x-1)) is -4/9 ln|x-1|+1/(3(x-1))+4/9 ln|x-4|+C

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integral of x/((x-1)(x-4)(x-1))

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

- \frac{4}{9} ln ∣ x - 1 ∣ + \frac{1}{3 ( x - 1 )} + \frac{4}{9} ln ∣ x - 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x - 1 ) ( x - 4 ) ( x - 1 )} : - \frac{4}{9 ( x - 1 )} - \frac{1}{3 ( x - 1 ) ^{2}} + \frac{4}{9 ( x - 4 )}

= \int - \frac{4}{9 ( x - 1 )} - \frac{1}{3 ( x - 1 ) ^{2}} + \frac{4}{9 ( x - 4 )} 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{4}{9 ( x - 1 )} d x - \int \frac{1}{3 ( x - 1 ) ^{2}} d x + \int \frac{4}{9 ( x - 4 )} d x

\int \frac{4}{9 ( x - 1 )} d x = \frac{4}{9} ln ∣ x - 1 ∣

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

\int \frac{4}{9 ( x - 4 )} d x = \frac{4}{9} ln ∣ x - 4 ∣

= - \frac{4}{9} ln ∣ x - 1 ∣ - (- \frac{1}{3 ( x - 1 )}) + \frac{4}{9} ln ∣ x - 4 ∣

Simplify

= - \frac{4}{9} ln ∣ x - 1 ∣ + \frac{1}{3 ( x - 1 )} + \frac{4}{9} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{4}{9} ln ∣ x - 1 ∣ + \frac{1}{3 ( x - 1 )} + \frac{4}{9} ln ∣ x - 4 ∣ + C

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What is the integral of x/((x-1)(x-4)(x-1)) ?
The integral of x/((x-1)(x-4)(x-1)) is -4/9 ln|x-1|+1/(3(x-1))+4/9 ln|x-4|+C