What is the integral of 1/(x(x^2-81)^{3/2)} ?

The integral of 1/(x(x^2-81)^{3/2)} is 1/729 (-arcsec(1/9 x)-9/((x^2-81)^{1/2)})+C

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integral of 1/(x(x^2-81)^{3/2)}

\int \frac{1}{x ( x ^{2} - 81 ) ^{\frac{3}{2}}} d x

\frac{1}{729} (- arcsec (\frac{1}{9} x) - \frac{9}{( x ^{2} - 81 ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{1}{x ( x ^{2} - 81 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{729 tan ^{2} ( u )} d u

= \int \frac{1}{729 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{729} \cdot \int \frac{1}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{729} \cdot \int - 1 + csc^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{729} (- \int 1 d u + \int csc^{2} (u) d u)

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{729} (- u - cot (u))

Substitute back

u = arcsec (\frac{1}{9} x)

= \frac{1}{729} (- arcsec (\frac{1}{9} x) - cot (arcsec (\frac{1}{9} x)))

Simplify

\frac{1}{729} (- arcsec (\frac{1}{9} x) - cot (arcsec (\frac{1}{9} x))) : \frac{1}{729} (- arcsec (\frac{1}{9} x) - \frac{9}{( x ^{2} - 81 ) ^{\frac{1}{2}}})

= \frac{1}{729} (- arcsec (\frac{1}{9} x) - \frac{9}{( x ^{2} - 81 ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{729} (- arcsec (\frac{1}{9} x) - \frac{9}{( x ^{2} - 81 ) ^{\frac{1}{2}}}) + C

y = \frac{( x + 2 )}{( x + 1 ) ^{2}}, at x = 0

\frac{\partial}{\partial x} (sin (π (5 x - y)))

y^{2} y^{^{'}} = e^{x}

x \to \infty lim (f (x))

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

What is the integral of 1/(x(x^2-81)^{3/2)} ?
The integral of 1/(x(x^2-81)^{3/2)} is 1/729 (-arcsec(1/9 x)-9/((x^2-81)^{1/2)})+C