What is the integral of 1/(x^5sqrt(x^2-1)) ?

The integral of 1/(x^5sqrt(x^2-1)) is (sqrt(x^2-1))/(x^4)+3/8 (arcsec(x)-1/4 sin(4arcsec(x)))+C

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integral of 1/(x^5sqrt(x^2-1))

\int \frac{1}{x ^{5} x ^{2} - 1} d x

\frac{x ^{2} - 1}{x ^{4}} + \frac{3}{8} (arcsec (x) - \frac{1}{4} sin (4 arcsec (x))) + C

Solution steps

\int \frac{1}{x ^{5} x ^{2} - 1} d x

Apply Trigonometric Substitution

= \int cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

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

Apply Integration By Parts

= cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u)))

Substitute back

u = arcsec (x)

= cos^{3} (arcsec (x)) sin (arcsec (x)) - (- \frac{3}{8} (arcsec (x) - \frac{1}{4} sin (4 arcsec (x))))

Simplify

cos^{3} (arcsec (x)) sin (arcsec (x)) - (- \frac{3}{8} (arcsec (x) - \frac{1}{4} sin (4 arcsec (x)))) : \frac{x ^{2} - 1}{x ^{4}} + \frac{3}{8} (arcsec (x) - \frac{1}{4} sin (4 arcsec (x)))

= \frac{x ^{2} - 1}{x ^{4}} + \frac{3}{8} (arcsec (x) - \frac{1}{4} sin (4 arcsec (x)))

Add a constant to the solution

= \frac{x ^{2} - 1}{x ^{4}} + \frac{3}{8} (arcsec (x) - \frac{1}{4} sin (4 arcsec (x))) + C

d er i v a t i v e x^{2} - 3 x

\frac{d}{d x} (- \frac{( a x + b - d ) ( x ^{2} - 1 )}{x ^{2}})

\frac{d}{d x} (arccsc (\frac{x}{2}))

\frac{d}{d x} (\frac{4 + x}{4 - x})

d er i v a t i v e e^{\frac{1}{x}}

What is the integral of 1/(x^5sqrt(x^2-1)) ?
The integral of 1/(x^5sqrt(x^2-1)) is (sqrt(x^2-1))/(x^4)+3/8 (arcsec(x)-1/4 sin(4arcsec(x)))+C