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.