What is the integral of 1/((x^2-5)^{3/2)} ?

The integral of 1/((x^2-5)^{3/2)} is -x/(5(x^2-5)^{1/2)}+C

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integral of 1/((x^2-5)^{3/2)}

\int \frac{1}{( x ^{2} - 5 ) ^{\frac{3}{2}}} d x

- \frac{x}{5 ( x ^{2} - 5 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 5 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{5 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}{5} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{5} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{5} (- \frac{1}{v})

Substitute back

= \frac{1}{5} - \frac{1}{sin ( arcsec ( \frac{1}{5 ^{\frac{1}{2}}} x ) )}

Simplify

\frac{1}{5} - \frac{1}{sin ( arcsec ( \frac{1}{5 ^{\frac{1}{2}}} x ) )} : - \frac{x}{5 ( x ^{2} - 5 ) ^{\frac{1}{2}}}

= - \frac{x}{5 ( x ^{2} - 5 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x}{5 ( x ^{2} - 5 ) ^{\frac{1}{2}}} + C

tan (2) (x + 1)

\frac{d y}{d x} - 8 y = 5

\frac{\partial}{\partial x} (\frac{1}{x cos ( z )})

\int \frac{3 x}{2 e ^{6 x}} d x

\int_{0}^{5} \frac{1}{x ^{\frac{15}{14}}} d x

What is the integral of 1/((x^2-5)^{3/2)} ?
The integral of 1/((x^2-5)^{3/2)} is -x/(5(x^2-5)^{1/2)}+C