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

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

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

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

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

Solution steps

\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

Apply Trigonometric Substitution

= \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

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

Apply the Power Rule

= \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{1}{2}}}

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

Add a constant to the solution

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

\int (\frac{sin ( 31 x )}{1 + cos ^{2} ( 31 x )}) d x

d er i v a t i v e e^{t + 7}

x \to - 7 + lim (\frac{- 6 x}{x + 7})

d er i v a t i v e f (θ) = 3 θ

\frac{d}{d x} (\frac{8 x y z}{( x ^{2} + 1 ) ^{2}})

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