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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} u ^{2} + 2} d u

= \int \frac{2}{u ^{2} u ^{2} + 2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{u ^{2} u ^{2} + 2} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{w}

= 2 \cdot \frac{1}{2} (- \frac{1}{w})

Substitute back

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

Simplify

2 \cdot \frac{1}{2} - \frac{1}{sin ( arctan ( \frac{1}{2} x - 2 ) )} : - \frac{x}{x - 2}

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

Add a constant to the solution

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

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