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

The integral of 1/(xsqrt(x^2-a^2)) is 1/a arctan((sqrt(x^2-a^2))/a)+C

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

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

\frac{1}{a} arctan (\frac{x ^{2} - a ^{2}}{a}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u u - a ^{2}} d u

= \int \frac{1}{2 u u - a ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u u - a ^{2}} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + a ^{2}} d v

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + a ^{2}} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + a ^{2}} d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{a ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{a} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{2} \cdot 2 \cdot \frac{1}{a} arctan (w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{a} arctan (\frac{x ^{2} - a ^{2}}{a})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{a} arctan (\frac{x ^{2} - a ^{2}}{a}) : \frac{1}{a} arctan (\frac{x ^{2} - a ^{2}}{a})

= \frac{1}{a} arctan (\frac{x ^{2} - a ^{2}}{a})

Add a constant to the solution

= \frac{1}{a} arctan (\frac{x ^{2} - a ^{2}}{a}) + C

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What is the integral of 1/(xsqrt(x^2-a^2)) ?
The integral of 1/(xsqrt(x^2-a^2)) is 1/a arctan((sqrt(x^2-a^2))/a)+C