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

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

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integral of 1/(sqrt(x^2\pm a^2))

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

\frac{1}{\pm a ^{2}} ln ∣ x ∣ + C

Solution steps

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

\frac{1}{x ^{2} \pm a ^{2}} = \frac{1}{\pm a ^{2}} \frac{1}{x ^{2}}

= \int \frac{1}{\pm a ^{2}} \cdot \frac{1}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{\pm a ^{2}} \cdot \int \frac{1}{x ^{2}} d x

x^{2} = (x),

assuming

x \geq 0

= \frac{1}{\pm a ^{2}} \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{\pm a ^{2}} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{\pm a ^{2}} ln ∣ x ∣ + C

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