How do you find the integral of $[dx / ((4-x)^2)^(3/2)]$ ?

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Answer 1 · 2015-07-29T12:45:22

Procedure is outlined below.

We have, $int dx/[(4 - x)^2]^(3/2)$

The denominator can be simply written as,

$[(4 - x)^2]^(3/2)$ = $(4 - x)^3$ by properties of exponents.

Thus, the integral now becomes,

$int dx/(4 - x)^3$

We now substitute, $(4 - x) = t$
$implies dx = dt$

Therefore, $int dx/(4 - x)^3 = int dt/t^3$
$= t^(-3+1)/(-3+1) + C$

Thus, in terms of $x$ ,

$int dx/[(4 - x)^2]^(3/2) = - (4 - x)^-2/2 +C$

Here, $C$ is the constant of integration.

How do you find the integral of [dx / ((4-x)^2)^(3/2)] [dx / ((4-x)^2)^(3/2)]?