What is the integral of 4/((sqrt(4x^2-1))^3) ?

The integral of 4/((sqrt(4x^2-1))^3) is -(4x)/(sqrt(4x^2-1))+C

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integral of 4/((sqrt(4x^2-1))^3)

\int \frac{4}{( 4 x ^{2} - 1 ) ^{3}} d x

- \frac{4 x}{4 x ^{2} - 1} + C

Solution steps

\int \frac{4}{( 4 x ^{2} - 1 ) ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{( 4 x ^{2} - 1 ) ^{3}} d x

(4 x^{2} - 1)^{3} = (4 x^{2} - 1)^{\frac{3}{2}}

= 4 \cdot \int \frac{1}{( 4 x ^{2} - 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 4 \cdot \int \frac{sec ( u )}{2 tan ^{2} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 4 \cdot \frac{1}{2} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 4 \cdot \frac{1}{2} (- \frac{1}{v})

Substitute back

= 4 \cdot \frac{1}{2} (- \frac{1}{sin ( arcsec ( 2 x ) )})

Simplify

4 \cdot \frac{1}{2} (- \frac{1}{sin ( arcsec ( 2 x ) )}) : - \frac{4 x}{4 x ^{2} - 1}

= - \frac{4 x}{4 x ^{2} - 1}

Add a constant to the solution

= - \frac{4 x}{4 x ^{2} - 1} + C

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What is the integral of 4/((sqrt(4x^2-1))^3) ?
The integral of 4/((sqrt(4x^2-1))^3) is -(4x)/(sqrt(4x^2-1))+C