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

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

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

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

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

Solution steps

\int \frac{1}{x ^{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 ^{2}} d x - \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

x \to 0 lim (\frac{5 sin ( x )}{x})

\int (x - 1)^{2} (x - 2) (x - 4) d x

x^{2} - 7 x

\frac{\partial}{\partial x} (\frac{7 x}{4 + x ^{2} + y ^{2}})

(lo g_{10} (1 + x))^{^{'}}