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

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

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

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

2 ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

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

Expand

\frac{2 x - 1}{x ^{2}} : \frac{2}{x} - \frac{1}{x ^{2}}

= \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

3 x^{3} - x^{2} + 9 x - 9

\frac{s - 2}{s ^{2} - s - 6}

\frac{d}{d x} (\frac{arcsin ( 3 x )}{x})

f (x) = 3 - 4 x^{2}, at x = - 2

y = 3 x, (3, 3)

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