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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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

assuming

(x^{2} - 4 x) \geq 0

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

Complete the square

x^{2} - 4 x : (x - 2)^{2} - 4

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

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 ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{w}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\int x 64 - 16 x^{2} d x

\int \frac{csc ^{2} ( 2 x )}{x} d x

\int \frac{2 x + 1}{3 x + 2} d x

\int x^{3} e^{- a x} d x

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