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

\frac{1}{x + 4}

\int \frac{20}{( x - 1 ) ( x ^{2} + 9 )} d x

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

(x^{2} + 2 x) (1 - e^{x})

\frac{d}{d x} (\frac{x + 4}{x ^{3} + x - 5})

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