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

The integral of ((x^2-a^2)^{1/2})/x is -aarctan(((x^2-a^2)^{1/2})/a)+(x^2-a^2)^{1/2}+C

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integral of ((x^2-a^2)^{1/2})/x

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d v = v

= a (- arctan (v) + v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of ((x^2-a^2)^{1/2})/x ?
The integral of ((x^2-a^2)^{1/2})/x is -aarctan(((x^2-a^2)^{1/2})/a)+(x^2-a^2)^{1/2}+C