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

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

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

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

- \frac{1}{x} + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - \frac{1}{x} + ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{x} + ln ∣ x ∣ + C

f (x) = - x^{2} + 3 x + 2

f (x) = 5 x^{2} + 3 x + 8

\frac{7}{8 x ^{3}}

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

f (x) = \frac{1}{( 1 + x ^{2} )}, at x = 1

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