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

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

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

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

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

Solution steps

\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

\int x e^{- x^{4}} d x

\int \frac{( x ^{3} )}{x + 1} d x

\int \frac{cos ( ln ( x ) )}{6 x} d x

\int x tan (x) sec^{2} (x) d x

\int 5 cos (θ) d θ

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