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

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

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

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

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

Solution steps

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

Simplify

\frac{x}{x ^{4} x ^{4} - x ^{2}} : \frac{1}{x ^{4} x ^{2} - 1}

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

Apply Trigonometric Substitution

= \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

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

d er i v a t i v e f (x) = (x + 1) (x + 1 - \frac{1}{x + 2})

in v erse l a pl a ce \frac{s - 1}{s ( s + 2.5 )}

\frac{\partial}{\partial x} (3 x z e^{4 x y})

x \to - 3 lim (3 x^{2} - 4 x + 1)

(x + 1) (\frac{d y}{d x}) + (2 x - 1) y = e^{- 2 x}

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