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

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

\int \frac{1}{x ^{5} \cdot 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

= \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

= 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

\frac{\partial}{\partial x} (4 x y)

3 t^{2} + 2 t - 2

\frac{6 x ^{2} + 6 x + 4}{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 1

x \to 3 lim (cos (π x) + 6)

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