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

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

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integral of (16-x^2)/x

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

16 ln ∣ x ∣ - \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

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

\int \frac{16}{x} d x = 16 ln ∣ x ∣

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

= 16 ln ∣ x ∣ - \frac{x ^{2}}{2}

Add a constant to the solution

= 16 ln ∣ x ∣ - \frac{x ^{2}}{2} + C

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