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

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

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

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

ln ∣ x ∣ + C

Solution steps

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

x^{2} = (x),

assuming

x \geq 0

= \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= ln ∣ x ∣

Add a constant to the solution

= ln ∣ x ∣ + C

x \to 1 lim (5 - 2 x)

x \to - π lim (x + 4)

\frac{d}{d x} (lo g_{3} (\frac{x x - 1}{2}))

\int \frac{10}{x ^{2} x ^{2} + 9} d x

y = 565 + t (37 t^{0.6} - 103)

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