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

The integral of 1/((x-7)^2)-1 is -1/(x-7)-x+C

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

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

- \frac{1}{x - 7} - x + C

Solution steps

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

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

\int 1 d x = x

= - \frac{1}{x - 7} - x

Add a constant to the solution

= - \frac{1}{x - 7} - x + C

n = 1 \sum \infty \frac{1}{2 n}

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

x \to 0 lim (6 x + ∣ x - 5 ∣)

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 8}{x})

f (x) = lo g_{2} (x)

What is the integral of 1/((x-7)^2)-1 ?
The integral of 1/((x-7)^2)-1 is -1/(x-7)-x+C