What is the integral of 1/(sqrt(x^2)-4) ?

The integral of 1/(sqrt(x^2)-4) is ln|x-4|+C

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integral of 1/(sqrt(x^2)-4)

\int \frac{1}{x ^{2} - 4} d x

ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 4} d x

x^{2} = (x),

assuming

x \geq 0

= \int \frac{1}{x - 4} d x

Apply u-substitution

: \int \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x - 4

= ln ∣ x - 4 ∣

Add a constant to the solution

= ln ∣ x - 4 ∣ + C

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What is the integral of 1/(sqrt(x^2)-4) ?
The integral of 1/(sqrt(x^2)-4) is ln|x-4|+C