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

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

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

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

ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

\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})

Simplify

= ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= ln ∣ x ∣ + \frac{1}{x} + C

x y^{^{'}} = y + 9 x^{2} sin (x), y (π) = 0

\frac{\partial}{\partial z} (\frac{x}{y} + \frac{y}{z})

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

s l o p e (1.4) (- 5)

\int 6 \frac{sin ( 2 x )}{sin ( x )} d x

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