What is the integral of ((sqrt(x^2-16)))/(x^3) ?

The integral of ((sqrt(x^2-16)))/(x^3) is -(sqrt(x^2-16))/(2x^2)+1/8 arcsec(1/4 x)+C

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integral of ((sqrt(x^2-16)))/(x^3)

\int \frac{( x ^{2} - 16 )}{x ^{3}} d x

- \frac{x ^{2} - 16}{2 x ^{2}} + \frac{1}{8} arcsec (\frac{1}{4} x) + C

Solution steps

\int \frac{x ^{2} - 16}{x ^{3}} d x

Apply Integration By Parts

: - \frac{x ^{2} - 16}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 16} d x

= - \frac{x ^{2} - 16}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 16} d x

\int - \frac{1}{2 x x ^{2} - 16} d x = - \frac{1}{8} arcsec (\frac{1}{4} x)

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

Simplify

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

Add a constant to the solution

= - \frac{x ^{2} - 16}{2 x ^{2}} + \frac{1}{8} arcsec (\frac{1}{4} x) + C

y^{^{'}} = 6 y + 2 t - 4, y (0) = 3

\int_{0}^{1} 1 - x d x

\frac{d}{d x} ((2 - x)^{\frac{3}{2}})

\frac{\partial}{\partial x} (4 x - 8 y^{3})

y^{^{''}} - 6 y = 0

What is the integral of ((sqrt(x^2-16)))/(x^3) ?
The integral of ((sqrt(x^2-16)))/(x^3) is -(sqrt(x^2-16))/(2x^2)+1/8 arcsec(1/4 x)+C