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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{81 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{81 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}{81} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{81} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{81} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{81} (- \frac{1}{v})

Substitute back

= \frac{1}{81} (- \frac{1}{sin ( arcsec ( \frac{1}{9} x ) )})

Simplify

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

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

Add a constant to the solution

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

\frac{3}{( s + 2 ) ^{3}}

y = x^{3} - 2 x^{2} + 2, y = 3 x^{2} + 5 x - 23, - 1, 7

\frac{\partial}{\partial x} (x^{2} e^{- \frac{x}{y}})

\int_{1}^{5} \frac{e ^{\frac{x}{2}}}{x ^{3}} d x

3 e^{x}, 3 x e^{x^{2}}, [0, 1]

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