How do you use substitution to integrate # x/sqrt(x+4) dx#?

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36
Jul 29, 2015

Answer:

The process is outlined below.

Explanation:

We are to evaluate, #int x/(x + 4)^(1/2)*dx#.

We shall accomplish this by substitution.

Let, #(x + 4) = t#
#implies dx = dt# (By simple differentiation).

Thus, the integral becomes,

#int x/(x +4)^(1/2)*dx = int (t - 4)/t^(1/2)*dt#

# = int t^(1/2)*dt - int 4t^(-1/2)*dt#

# =2/3 t^(3/2) - 8t^(1/2) + C#, where #C# is the integration constant.

In terms of #x#, the integral may be now written as,

#int x/(x + 4)^(1/2)*dx =2/3 (x + 4)^(3/2) - 8(x + 4)^(1/2) + C#.

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Aug 15, 2017

Answer:

#int x/(sqrt(x+4)) "d"x = 2/3 (x+4)sqrt(x+4) - 8 sqrt(x+4) + C#

Explanation:

#int x/(sqrt(x+4)) "d"x = int (x+4-4)/(sqrt(x+4)) "d"x#,
#int x/(sqrt(x+4)) "d"x = int (x+4)/((x+4)^(1/2))-4/((x+4)^(1/2)) "d"x#,
#int x/(sqrt(x+4)) "d"x = int (x+4)^(1/2)-4(x+4)^(-1/2) "d"x#,
#int x/(sqrt(x+4)) "d"x = 2/3(x+4)^(3/2)-8(x+4)^(1/2) + C#

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5
Aug 11, 2017

Let #u^2 = x+4#
Then #2u*du = dx#
#x = u^2-4#
Now the integral becomes
#int (u^2-4)/u * 2u*du#
= #2int (u^2-4) du#
= #2(u^3/3 - 4u)#
= #2/3(x+4)^(3/2) - 8sqrt(x+4)#

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