What is the integral of 1/((v^2-3)^{3/2)} ?

The integral of 1/((v^2-3)^{3/2)} is -v/(3(v^2-3)^{1/2)}+C

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integral of 1/((v^2-3)^{3/2)}

\int \frac{1}{( v ^{2} - 3 ) ^{\frac{3}{2}}} d v

- \frac{v}{3 ( v ^{2} - 3 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( v ^{2} - 3 ) ^{\frac{3}{2}}} d v

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{3 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{3} (- \frac{1}{w})

Substitute back

= \frac{1}{3} - \frac{1}{sin ( arcsec ( \frac{1}{3 ^{\frac{1}{2}}} v ) )}

Simplify

\frac{1}{3} - \frac{1}{sin ( arcsec ( \frac{1}{3 ^{\frac{1}{2}}} v ) )} : - \frac{v}{3 ( v ^{2} - 3 ) ^{\frac{1}{2}}}

= - \frac{v}{3 ( v ^{2} - 3 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{v}{3 ( v ^{2} - 3 ) ^{\frac{1}{2}}} + C

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What is the integral of 1/((v^2-3)^{3/2)} ?
The integral of 1/((v^2-3)^{3/2)} is -v/(3(v^2-3)^{1/2)}+C