What is the integral of 1/(x^5sqrt(4x^2-1)) ?

The integral of 1/(x^5sqrt(4x^2-1)) is (sqrt(4x^2-1))/(x^4)+6arcsec(2x)-3/2 sin(4arcsec(2x))+C

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integral of 1/(x^5sqrt(4x^2-1))

\int \frac{1}{x ^{5} 4 x ^{2} - 1} d x

\frac{4 x ^{2} - 1}{x ^{4}} + 6 arcsec (2 x) - \frac{3}{2} sin (4 arcsec (2 x)) + C

Solution steps

\int \frac{1}{x ^{5} 4 x ^{2} - 1} d x

Apply Trigonometric Substitution

: \int 16 cos^{4} (u) d u

= \int 16 cos^{4} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int cos^{4} (u) d u

Simplify

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

= 16 \cdot \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

= 16 (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))

= 16 (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = arcsec (2 x)

= 16 (cos^{3} (arcsec (2 x)) sin (arcsec (2 x)) - (- \frac{3}{8} (arcsec (2 x) - \frac{1}{4} sin (4 arcsec (2 x)))))

Simplify

16 (cos^{3} (arcsec (2 x)) sin (arcsec (2 x)) - (- \frac{3}{8} (arcsec (2 x) - \frac{1}{4} sin (4 arcsec (2 x))))) : \frac{4 x ^{2} - 1}{x ^{4}} + 6 arcsec (2 x) - \frac{3}{2} sin (4 arcsec (2 x))

= \frac{4 x ^{2} - 1}{x ^{4}} + 6 arcsec (2 x) - \frac{3}{2} sin (4 arcsec (2 x))

Add a constant to the solution

= \frac{4 x ^{2} - 1}{x ^{4}} + 6 arcsec (2 x) - \frac{3}{2} sin (4 arcsec (2 x)) + C

\int \frac{x}{sin ^{2} ( x )} d x

\frac{d}{d x} (arcsin (x))

1 - e^{x}

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{4 + x ^{2} y ^{2}})

x \to 0 + lim (\frac{e ^{- \frac{1}{x}}}{x})

What is the integral of 1/(x^5sqrt(4x^2-1)) ?
The integral of 1/(x^5sqrt(4x^2-1)) is (sqrt(4x^2-1))/(x^4)+6arcsec(2x)-3/2 sin(4arcsec(2x))+C