What is the integral of (x^2-16)/(x^2) ?

The integral of (x^2-16)/(x^2) is x+(16)/x+C

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integral of (x^2-16)/(x^2)

\int \frac{x ^{2} - 16}{x ^{2}} d x

x + \frac{16}{x} + C

Solution steps

\int \frac{x ^{2} - 16}{x ^{2}} d x

Expand

\frac{x ^{2} - 16}{x ^{2}} : 1 - \frac{16}{x ^{2}}

= \int 1 - \frac{16}{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{16}{x ^{2}} d x

\int 1 d x = x

\int \frac{16}{x ^{2}} d x = - \frac{16}{x}

= x - (- \frac{16}{x})

Simplify

= x + \frac{16}{x}

Add a constant to the solution

= x + \frac{16}{x} + C

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What is the integral of (x^2-16)/(x^2) ?
The integral of (x^2-16)/(x^2) is x+(16)/x+C