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

\frac{d}{d y} (ln (y) - \frac{y ^{2}}{8})

\int cos (4 x) d x

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

\frac{2}{s ^{2} + 3 s - 4}

\frac{1}{3} t - 2 (1 - e^{- \frac{t}{4}})

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