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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \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

: \int \frac{1}{v ^{2}} d v

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

Apply the Power Rule

: - \frac{1}{v}

= \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{1}{2}}}

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{1}{e ^{x} - 1})

f (x) = e^{x} \cdot x

\frac{d}{d x} (x^{y^{x}})

\int \frac{3}{x ^{2} + 4} d x

\int x^{4} x^{5} + 3 d x

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