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

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

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

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

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

Solution steps

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

Expand

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

= \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

y^{^{''}} + 4 y^{^{'}} + 20 y = 80

y = x, y = x - 2, y = 0

\int \frac{1}{( x ^{2} + 81 ) ^{2}} d x

\int 4 cos^{2} (θ) d θ

\int_{0}^{\frac{1}{2}} tan (\frac{x}{2}) d x

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