What is the integral of 1/(x^4sqrt(x^2-1)) ?

The integral of 1/(x^4sqrt(x^2-1)) is (sqrt(x^2-1))/x-((x^2-1)^{3/2})/(3x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^4sqrt(x^2-1))

\int \frac{1}{x ^{4} x ^{2} - 1} d x

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

Solution steps

\int \frac{1}{x ^{4} x ^{2} - 1} d x

Apply Trigonometric Substitution

: \int cos^{3} (u) d u

= \int cos^{3} (u) d u

Rewrite using trig identities

= \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d v - \int v^{2} d v

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{x} sin (x + y))

\frac{8}{x x ^{2} - 4}

x \to \infty lim ((k)^{x})

x \to 1 lim (tan (x))

y^{^{'}} = \frac{( x ^{4} y ^{4} + 4 x ^{4} )}{y ^{3}}

What is the integral of 1/(x^4sqrt(x^2-1)) ?
The integral of 1/(x^4sqrt(x^2-1)) is (sqrt(x^2-1))/x-((x^2-1)^{3/2})/(3x^3)+C