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

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

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integral of (x^2)/(sqrt((x^2-1)^5))

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

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

Solution steps

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

(x^{2} - 1)^{5} = (x^{2} - 1)^{\frac{5}{2}},

assuming

(x^{2} - 1) \geq 0

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

Apply Trigonometric Substitution

: \int cot^{4} (u) sec^{3} (u) d u

= \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{v ^{4}} d v

= \int \frac{1}{v ^{4}} d v

Apply the Power Rule

: - \frac{1}{3 v ^{3}}

= - \frac{1}{3 v ^{3}}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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