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

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

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

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

- \frac{1}{x} + ln ∣ x ∣ + C

Solution steps

\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

x \to \frac{π}{2} lim (9 tan (x))

\frac{d}{d x} (ln (ln (6 x)))

\int e^{t (1 - s)} d t

\int_{0}^{\frac{π}{2}} - 6 cos^{5} (x) d x

\frac{d}{d x} (x^{2} ln (\frac{1}{x}))

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