What is the integral of 2/x-1/(x^2) ?

The integral of 2/x-1/(x^2) is 2ln|x|+1/x+C

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integral of 2/x-1/(x^2)

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

2 ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= 2 ln ∣ x ∣ - (- \frac{1}{x})

Simplify

= 2 ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= 2 ln ∣ x ∣ + \frac{1}{x} + C

\int \frac{2 x}{4 x - 5} d x

\frac{d r}{d t} = \frac{r ^{2}}{t}

(- 3 x + 2)^{2}

\frac{d}{d x} (5^{3 x} + 7)

(\frac{x + 7}{x - 7})^{5}

What is the integral of 2/x-1/(x^2) ?
The integral of 2/x-1/(x^2) is 2ln|x|+1/x+C