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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{\frac{3}{2}}} d u

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

Apply the Power Rule

: - \frac{2}{u ^{\frac{1}{2}}}

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

Substitute back

u = x - 4

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

Add a constant to the solution

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

x \to 0 lim (x^{- \frac{5}{l n ( x )}})

\frac{\partial}{\partial x} (x^{2} + 3 y^{2})

f (x) = x^{2} + 8, (5, 33)

y = \frac{8}{x}, (2, 4)

x \to \infty lim (x^{3} + x)

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