What is the integral of (v^2)/((1-v^2)^{5/2)} ?

The integral of (v^2)/((1-v^2)^{5/2)} is (v^3)/(3(1-v^2)^{3/2)}+C

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integral of (v^2)/((1-v^2)^{5/2)}

\int \frac{v ^{2}}{( 1 - v ^{2} ) ^{\frac{5}{2}}} d v

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

Solution steps

\int \frac{v ^{2}}{( 1 - v ^{2} ) ^{\frac{5}{2}}} d v

Apply Trigonometric Substitution

: \int \frac{sin ^{2} ( u )}{cos ^{4} ( u )} d u

= \int \frac{sin ^{2} ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

: \int w^{2} d w

= \int w^{2} d w

Apply the Power Rule

: \frac{w ^{3}}{3}

= \frac{w ^{3}}{3}

Substitute back

= \frac{tan ^{3} ( arcsin ( v ) )}{3}

Simplify

\frac{tan ^{3} ( arcsin ( v ) )}{3} : \frac{v ^{3}}{3 ( 1 - v ^{2} ) ^{\frac{3}{2}}}

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

Add a constant to the solution

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

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What is the integral of (v^2)/((1-v^2)^{5/2)} ?
The integral of (v^2)/((1-v^2)^{5/2)} is (v^3)/(3(1-v^2)^{3/2)}+C