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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt((x^2-1)^3))

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

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

Solution steps

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

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

assuming

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

= \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{x}{x ^{2} - 1}

Add a constant to the solution

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

x \to 3 lim (4 x^{2})

\frac{\partial}{\partial x} (a cos (x) sin (y))

x \to 1 lim (e^{\frac{1}{1 - x}})

\frac{d}{d x} (2 x^{\frac{3}{4}})

\frac{\partial}{\partial y} (x y + z^{2})

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