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

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

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

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

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

Solution steps

\int \frac{1}{( 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

y = 4 x^{2} - 1, y = 1 - 4 x^{2}

y^{'} = y (2 - y)

x \to 0 lim (2 x - x^{2})

\frac{\partial}{\partial s} (s^{2} t)

x \to - 1 lim (arcsin (x))

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