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

The integral of 1/((x-1)^2)+C is -1/(x-1)+Cx+C

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

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

- \frac{1}{x - 1} + C x + C

Solution steps

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

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

\int C d x = C x

= - \frac{1}{x - 1} + C x

Add a constant to the solution

= - \frac{1}{x - 1} + C x + C

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