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

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

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

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

3 ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

Solution steps

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

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

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

= 3 ln ∣ x - 1 ∣ - \frac{1}{x - 1}

Add a constant to the solution

= 3 ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

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