What is the integral of 1/(sqrt((x^2-9)^3)) ?

The integral of 1/(sqrt((x^2-9)^3)) is -x/(9sqrt(x^2-9))+C

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integral of 1/(sqrt((x^2-9)^3))

\int \frac{1}{( x ^{2} - 9 ) ^{3}} d x

- \frac{x}{9 x ^{2} - 9} + C

Solution steps

\int \frac{1}{( x ^{2} - 9 ) ^{3}} d x

(x^{2} - 9)^{3} = (x^{2} - 9)^{\frac{3}{2}},

assuming

(x^{2} - 9) \geq 0

= \int \frac{1}{( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \int \frac{sec ( u )}{9 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}{9} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

= \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

= \frac{1}{9} (- \frac{1}{v})

Substitute back

= \frac{1}{9} (- \frac{1}{sin ( arcsec ( \frac{1}{3} x ) )})

Simplify

\frac{1}{9} (- \frac{1}{sin ( arcsec ( \frac{1}{3} x ) )}) : - \frac{x}{9 x ^{2} - 9}

= - \frac{x}{9 x ^{2} - 9}

Add a constant to the solution

= - \frac{x}{9 x ^{2} - 9} + C

What is the integral of 1/(sqrt((x^2-9)^3)) ?
The integral of 1/(sqrt((x^2-9)^3)) is -x/(9sqrt(x^2-9))+C