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

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

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

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

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

Solution steps

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

Expand

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

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

\int 1 d x = x

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

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

Simplify

= x + \frac{1}{x}

Add a constant to the solution

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

\int x^{2} e^{- 18 x} d x

(2 x^{4} + y^{4}) d y - x^{3} y d x = 0

f (x) = y

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

\frac{d}{d x} (arcsin (1))

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