What is the integral of 1/((x^2-64)^{3/2)} ?

The integral of 1/((x^2-64)^{3/2)} is -x/(64(x^2-64)^{1/2)}+C

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integral of 1/((x^2-64)^{3/2)}

\int \frac{1}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x

- \frac{x}{64 ( x ^{2} - 64 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \int \frac{sec ( u )}{64 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}{64} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{64} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

= \frac{1}{64} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

= \frac{1}{64} (- \frac{1}{v})

Substitute back

= \frac{1}{64} (- \frac{1}{sin ( arcsec ( \frac{1}{8} x ) )})

Simplify

\frac{1}{64} (- \frac{1}{sin ( arcsec ( \frac{1}{8} x ) )}) : - \frac{x}{64 ( x ^{2} - 64 ) ^{\frac{1}{2}}}

= - \frac{x}{64 ( x ^{2} - 64 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x}{64 ( x ^{2} - 64 ) ^{\frac{1}{2}}} + C

d er i v a t i v e y = \frac{1}{2 x}

x \to 2 lim (\frac{x + 7}{x ^{2} - 4})

d er i v a t i v e cos (arcsin (x + 3))

\int x (x - 4) d x

\frac{d}{d x} (arctan (e^{2 x}))

What is the integral of 1/((x^2-64)^{3/2)} ?
The integral of 1/((x^2-64)^{3/2)} is -x/(64(x^2-64)^{1/2)}+C