What is the integral of (9-x^2)/(x^2) ?

The integral of (9-x^2)/(x^2) is -9/x-x+C

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integral of (9-x^2)/(x^2)

\int \frac{9 - x ^{2}}{x ^{2}} d x

- \frac{9}{x} - x + C

Solution steps

\int \frac{9 - x ^{2}}{x ^{2}} d x

Expand

\frac{9 - x ^{2}}{x ^{2}} : \frac{9}{x ^{2}} - 1

= \int \frac{9}{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{9}{x ^{2}} d x - \int 1 d x

\int \frac{9}{x ^{2}} d x = - \frac{9}{x}

\int 1 d x = x

= - \frac{9}{x} - x

Add a constant to the solution

= - \frac{9}{x} - x + C

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

\int cos^{3} (u) d u

x \to π lim (cos (3 x))

\frac{d}{d x} (2 x 64 - x^{2})

x \to - 1 lim (- 1^{3})

What is the integral of (9-x^2)/(x^2) ?
The integral of (9-x^2)/(x^2) is -9/x-x+C