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

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

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

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

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

Solution steps

\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

= \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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{3} 5 x e^{x} d x

f (x) = 3 x^{6} (x^{4} - 7 x)

\int x^{2} - 25 d x

f (x) = (x + \frac{1}{3 x})^{2}

x \to \infty lim (e^{- 6 x} cos (x))

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