What is the integral of 1/([(ax)^2-b^2]^{3/2)} ?

The integral of 1/([(ax)^2-b^2]^{3/2)} is -x/(b^2(a^2x^2-b^2)^{1/2)}+C

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integral of 1/([(ax)^2-b^2]^{3/2)}

\int \frac{1}{[ ( a x ) ^{2} - b ^{2} ] ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{1}{[ ( a x ) ^{2} - b ^{2} ] ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{a ( u ^{2} - b ^{2} ) ^{\frac{3}{2}}} d u

= \int \frac{1}{a ( u ^{2} - b ^{2} ) ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{( u ^{2} - b ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{b ^{2} tan ^{2} ( v )} d v

= \frac{1}{a} \cdot \int \frac{sec ( v )}{b ^{2} tan ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{a} \cdot \frac{1}{b ^{2}} \cdot \int \frac{sec ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{a} \cdot \frac{1}{b ^{2}} \cdot \int \frac{cos ( v )}{sin ^{2} ( v )} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

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

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{a} \cdot \frac{1}{b ^{2}} (- \frac{1}{w})

Substitute back

= \frac{1}{a} \cdot \frac{1}{b ^{2}} (- \frac{1}{sin ( arcsec ( \frac{1}{b} a x ) )})

Simplify

\frac{1}{a} \cdot \frac{1}{b ^{2}} (- \frac{1}{sin ( arcsec ( \frac{1}{b} a x ) )}) : - \frac{x}{b ^{2} ( a ^{2} x ^{2} - b ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

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What is the integral of 1/([(ax)^2-b^2]^{3/2)} ?
The integral of 1/([(ax)^2-b^2]^{3/2)} is -x/(b^2(a^2x^2-b^2)^{1/2)}+C