What is the integral of 1/((x^2-8x)^{3/2)} ?

The integral of 1/((x^2-8x)^{3/2)} is -(x-4)/(16(x^2-8x)^{1/2)}+C

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integral of 1/((x^2-8x)^{3/2)}

\int \frac{1}{( x ^{2} - 8 x ) ^{\frac{3}{2}}} d x

- \frac{x - 4}{16 ( x ^{2} - 8 x ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 8 x ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} - 8 x : (x - 4)^{2} - 16

= \int \frac{1}{( ( x - 4 ) ^{2} - 16 ) ^{\frac{3}{2}}} d x

Apply u-substitution

= \int \frac{1}{( u ^{2} - 16 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

= \int \frac{sec ( v )}{16 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}{16} \cdot \int \frac{sec ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{16} \cdot \int \frac{cos ( v )}{sin ^{2} ( v )} d v

Apply u-substitution

= \frac{1}{16} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

= \frac{1}{16} (- \frac{1}{w})

Substitute back

= \frac{1}{16} (- \frac{1}{sin ( arcsec ( \frac{1}{4} ( x - 4 ) ) )})

Simplify

\frac{1}{16} (- \frac{1}{sin ( arcsec ( \frac{1}{4} ( x - 4 ) ) )}) : - \frac{x - 4}{16 ( x ^{2} - 8 x ) ^{\frac{1}{2}}}

= - \frac{x - 4}{16 ( x ^{2} - 8 x ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x - 4}{16 ( x ^{2} - 8 x ) ^{\frac{1}{2}}} + C

What is the integral of 1/((x^2-8x)^{3/2)} ?
The integral of 1/((x^2-8x)^{3/2)} is -(x-4)/(16(x^2-8x)^{1/2)}+C