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}

Add a constant to the solution

= 2 ln ∣ x ∣ - \frac{1}{x} + C

\frac{d}{d x} ((6 x)^{- 5})

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

y^{^{''}} + 25 y = t sin (5 t)

(x^{2} - x y) d x + x^{2} d y = 0

n = 1 \sum \infty \frac{11 n}{n !}

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