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

f (x) = (4 x - 1) cos (x)

f (x) = 2^{3} e^{2}

h \to + 0 lim (- 2 x - h)

\int (\frac{x + 1}{x ^{2} + 2 x}) d x

\frac{\partial}{\partial x} (x^{2} y - x y^{2})

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