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

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

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

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

ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{x} + C

\int \frac{1}{x ^{2} + 5 x + 6} d x

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

\frac{d}{d x} (ln (x + y - 1))

x^{2} (\frac{d y}{d x}) = y - y x

d er i v a t i v e y = (x + 2)^{2} - 3

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