What is the integral of 1/(usqrt(u^2-a^2)) ?

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

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integral of 1/(usqrt(u^2-a^2))

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

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

Solution steps

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

Apply u-substitution

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

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}{v - a ^{2} + v} d v

Apply u-substitution

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

Apply Integral Substitution

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

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}{t ^{2} + 1} d t

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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