What is the integral of (4/x-(16)/(x^2)) ?

The integral of (4/x-(16)/(x^2)) is 4ln|x|+(16)/x+C

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integral of (4/x-(16)/(x^2))

\int (\frac{4}{x} - \frac{16}{x ^{2}}) d x

4 ln ∣ x ∣ + \frac{16}{x} + C

Solution steps

\int \frac{4}{x} - \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 \frac{4}{x} d x - \int \frac{16}{x ^{2}} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

= 4 ln ∣ x ∣ - (- \frac{16}{x})

Simplify

= 4 ln ∣ x ∣ + \frac{16}{x}

Add a constant to the solution

= 4 ln ∣ x ∣ + \frac{16}{x} + C

\frac{\partial}{\partial y} (\frac{e ^{y}}{x + y ^{9}})

\frac{d}{d x} (2^{- 2 x})

f (x) = cos (x) - sin (x)

n = 0 \sum \infty \frac{2 n}{n} x^{n}

\frac{\partial}{\partial x} (e^{x} sin (y + z) - z)

What is the integral of (4/x-(16)/(x^2)) ?
The integral of (4/x-(16)/(x^2)) is 4ln|x|+(16)/x+C