What is the integral of (x^2-4)/(x^2) ?

The integral of (x^2-4)/(x^2) is x+4/x+C

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integral of (x^2-4)/(x^2)

\int \frac{x ^{2} - 4}{x ^{2}} d x

x + \frac{4}{x} + C

Solution steps

\int \frac{x ^{2} - 4}{x ^{2}} d x

Expand

\frac{x ^{2} - 4}{x ^{2}} : 1 - \frac{4}{x ^{2}}

= \int 1 - \frac{4}{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{4}{x ^{2}} d x

\int 1 d x = x

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

= x - (- \frac{4}{x})

Simplify

= x + \frac{4}{x}

Add a constant to the solution

= x + \frac{4}{x} + C

x \to 1 lim (\frac{x - 2}{x ^{2} - 4})

y = \frac{4 x - 9}{2 x + 3}, x = 2

(2 x - 2)^{4} (x^{2} + x + 1)^{5}

\int_{0}^{4} x e^{x} d x

(ln (x^{2} + 1))^{^{'}}

What is the integral of (x^2-4)/(x^2) ?
The integral of (x^2-4)/(x^2) is x+4/x+C