What is the integral of 5/x-(10)/(x^2) ?

The integral of 5/x-(10)/(x^2) is 5ln|x|+(10)/x+C

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integral of 5/x-(10)/(x^2)

\int \frac{5}{x} - \frac{10}{x ^{2}} d x

5 ln ∣ x ∣ + \frac{10}{x} + C

Solution steps

\int \frac{5}{x} - \frac{10}{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{5}{x} d x - \int \frac{10}{x ^{2}} d x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

\int \frac{10}{x ^{2}} d x = - \frac{10}{x}

= 5 ln ∣ x ∣ - (- \frac{10}{x})

Simplify

= 5 ln ∣ x ∣ + \frac{10}{x}

Add a constant to the solution

= 5 ln ∣ x ∣ + \frac{10}{x} + C

\frac{d}{d x} ((x^{3} - 2 x + 1) (3 x^{3} + 2 x^{2} - 5 x))

x \to - \infty lim (\frac{x}{- 1})

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

f (x) = (x - 4) (x^{2} + 8)

y^{^{'}} = 4 y + y^{4}

What is the integral of 5/x-(10)/(x^2) ?
The integral of 5/x-(10)/(x^2) is 5ln|x|+(10)/x+C