What is the integral of 1/(sqrt((x^2-a^2)^3)) ?

The integral of 1/(sqrt((x^2-a^2)^3)) is -x/(a^2sqrt(x^2-a^2))+C

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integral of 1/(sqrt((x^2-a^2)^3))

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

- \frac{x}{a ^{2} x ^{2} - a ^{2}} + C

Solution steps

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

(x^{2} - a^{2})^{3} = (x^{2} - a^{2})^{\frac{3}{2}},

assuming

(x^{2} - a^{2}) \geq 0

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{a ^{2}} (- \frac{1}{v})

Substitute back

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

Simplify

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

= - \frac{x}{a ^{2} x ^{2} - a ^{2}}

Add a constant to the solution

= - \frac{x}{a ^{2} x ^{2} - a ^{2}} + C

\int - 0.08 x^{2} + 4 x d x

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What is the integral of 1/(sqrt((x^2-a^2)^3)) ?
The integral of 1/(sqrt((x^2-a^2)^3)) is -x/(a^2sqrt(x^2-a^2))+C