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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec^{4} (u) d u

= \int sec^{4} (u) d u

Rewrite using trig identities

= \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int 1 + v^{2} d v

= \int 1 + v^{2} 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

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

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= v + \frac{v ^{3}}{3}

Substitute back

= tan (arcsin (x)) + \frac{tan ^{3} ( arcsin ( x ) )}{3}

Simplify

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

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

Add a constant to the solution

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

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