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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int sin (8 x + 2) d x

2 \cdot y^{^{''}} + y^{^{'}} - y = 0

y^{^{''}} + 2 y^{^{'}} = 2 t + 5 - e^{- 2 t}

\frac{d}{d x} (x^{2 - 2 x})

y = \frac{ln ( x )}{e ^{x}}

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