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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{2}{u ^{2} + a ^{2}} d u

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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