What is the integral of (sqrt(x^2-36))/(x^3) ?

The integral of (sqrt(x^2-36))/(x^3) is -(sqrt(x^2-36))/(2x^2)+1/12 arcsec(1/6 x)+C

Popular >

integral of (sqrt(x^2-36))/(x^3)

\int \frac{x ^{2} - 36}{x ^{3}} d x

- \frac{x ^{2} - 36}{2 x ^{2}} + \frac{1}{12} arcsec (\frac{1}{6} x) + C

Solution steps

\int \frac{x ^{2} - 36}{x ^{3}} d x

Apply Integration By Parts

: - \frac{x ^{2} - 36}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 36} d x

= - \frac{x ^{2} - 36}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 36} d x

\int - \frac{1}{2 x x ^{2} - 36} d x = - \frac{1}{12} arcsec (\frac{1}{6} x)

= - \frac{x ^{2} - 36}{2 x ^{2}} - (- \frac{1}{12} arcsec (\frac{1}{6} x))

Simplify

= - \frac{x ^{2} - 36}{2 x ^{2}} + \frac{1}{12} arcsec (\frac{1}{6} x)

Add a constant to the solution

= - \frac{x ^{2} - 36}{2 x ^{2}} + \frac{1}{12} arcsec (\frac{1}{6} x) + C

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 11

\int 5 tan (2 x) ln (cos (2 x)) d x

\frac{\partial}{\partial t} (z y^{2} (t + 1))

\frac{d y}{d x} = \frac{x y ^{3}}{1 + x ^{2}}, y (0) = - 1

\frac{d}{d x} (sin (3 x) - 4 x^{4} + 2 x^{3} + 3)

What is the integral of (sqrt(x^2-36))/(x^3) ?
The integral of (sqrt(x^2-36))/(x^3) is -(sqrt(x^2-36))/(2x^2)+1/12 arcsec(1/6 x)+C