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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cot^{2} (u) sec (u) d u

= \int cot^{2} (u) sec (u) d u

Simplify

cot^{2} (u) sec (u) : cot (u) csc (u)

= \int cot (u) csc (u) d u

Apply u-substitution

: \int - 1 d v

= \int - 1 d v

Integral of a constant:

\int a d x = a x

= (- 1) v

Substitute back

= (- 1) csc (arcsec (x))

Simplify

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

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

Add a constant to the solution

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

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\int \frac{1}{z z ^{2} - 1} d z

\int \frac{x ^{7} - 2 x}{x ^{5}} d x

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

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