What is the integral of 1/((x-5)^2)-1 ?

The integral of 1/((x-5)^2)-1 is -1/(x-5)-x+C

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integral of 1/((x-5)^2)-1

\int \frac{1}{( x - 5 ) ^{2}} - 1 d x

- \frac{1}{x - 5} - x + C

Solution steps

\int \frac{1}{( x - 5 ) ^{2}} - 1 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 - 5 ) ^{2}} d x - \int 1 d x

\int \frac{1}{( x - 5 ) ^{2}} d x = - \frac{1}{x - 5}

\int 1 d x = x

= - \frac{1}{x - 5} - x

Add a constant to the solution

= - \frac{1}{x - 5} - x + C

\frac{1}{2 x}

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What is the integral of 1/((x-5)^2)-1 ?
The integral of 1/((x-5)^2)-1 is -1/(x-5)-x+C