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

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

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

integral

\frac{x}{( x - 2 ) ^{\frac{3}{2}}}

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 2}{u ^{\frac{3}{2}}} d u

= \int \frac{u + 2}{u ^{\frac{3}{2}}} d u

Expand

\frac{u + 2}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} + \frac{2}{u ^{\frac{3}{2}}}

= \int \frac{1}{u ^{\frac{1}{2}}} + \frac{2}{u ^{\frac{3}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{\frac{1}{2}}} d u + \int \frac{2}{u ^{\frac{3}{2}}} d u

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

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

= 2 u^{\frac{1}{2}} - \frac{4}{u ^{\frac{1}{2}}}

Substitute back

u = x - 2

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

Add a constant to the solution

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

(e^{x})^{4}

\frac{1}{( x ^{2} + x )}

\frac{x}{( 5 x + 2 ) ^{3}}

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

\frac{1}{( a ^{2} - x ^{2} ) ^{\frac{1}{2}}}

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