What is the integral of 1/((x^2-36)^{3/2)} ?

The integral of 1/((x^2-36)^{3/2)} is -x/(36(x^2-36)^{1/2)}+C

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integral of 1/((x^2-36)^{3/2)}

\int \frac{1}{( x ^{2} - 36 ) ^{\frac{3}{2}}} d x

- \frac{x}{36 ( x ^{2} - 36 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 36 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{36 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{36 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}{36} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{36} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{36} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{36} (- \frac{1}{v})

Substitute back

= \frac{1}{36} (- \frac{1}{sin ( arcsec ( \frac{1}{6} x ) )})

Simplify

\frac{1}{36} (- \frac{1}{sin ( arcsec ( \frac{1}{6} x ) )}) : - \frac{x}{36 ( x ^{2} - 36 ) ^{\frac{1}{2}}}

= - \frac{x}{36 ( x ^{2} - 36 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x}{36 ( x ^{2} - 36 ) ^{\frac{1}{2}}} + C

\int \frac{1 - x ^{2}}{2} d x

csc^{2} (x) - 5

\frac{s - 4}{2 s ^{2} + s + 8}

\frac{\partial}{\partial y} (3 x^{2} + 5 y^{2})

x y^{^{'}} + y = - 2 x y^{2}, y (1) = 7

What is the integral of 1/((x^2-36)^{3/2)} ?
The integral of 1/((x^2-36)^{3/2)} is -x/(36(x^2-36)^{1/2)}+C