# Solve the indefinite integral `int 3x*(4x-1)^(1/2) dx` using the substitution u=4x-1

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

`int 3x(4x-1)^(1/2)dx` :

Let `u=4x-1,du=4,x=(u+1)/4` ; rewrite as

`int 3x*1/4*4(4x-1)^(1/2)dx` then pulling out constants and substituting we get:

`3/16 int (u+1)u^(1/2)du`

`=3/16int(u^(3/2)+u^(1/2))du`

`=3/16intu^(3/2)du+3/16intu^(1/2)du`

`=(3/16)(2/5)u^(5/2)+(3/16)(2/3)u^(3/2)+C`

`=3/40u^(5/2)+1/8u^(3/2)+C`

Factor out a common `u^(3/2)` :

`=u^(3/2)(3/40u+1/8)+C` Back substitute for u:

`=(4x-1)^(3/2)[3/40(4x-1)+1/8]+C`

`=(4x-1)^(3/2)[3/10x+1/20]+C`

`=1/20(4x-1)^(3/2)(6x+1)+C`

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The solution is:

`int3x(4x-1)^(1/2)dx=1/20(4x-1)^(3/2)(6x+1)+C`

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Check: Take the derivative of the solution to get:

`d/(dx)[1/20(4x-1)^(3/2)(6x+1)+C]`

`=1/20[6(4x-1)^(3/2)+(3/2)(4x-1)^(1/2)(4)(6x+1)]`

`=1/20[6(4x-1)^(3/2)+6(6x+1)(4x-1)^(1/2)]`

Factor out a common `6(4x-1)^(1/2)` to get:

`3/10(4x-1)^(1/2)[4x-1+6x+1]`

=`3/10(4x-1)^(1/2)[10x]`

`=3x(4x-1)^(1/2)` as required.

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

If you use U^2 = (4x-1) then U= (4x-1)^(1/2)

Then 2U*du = 4dx

(U/2)*du = dx

Also x = (U^2+1)/4

So;

∫3x(4x-1)^(1/2)dx = ∫[3*(U^2+1)/4]*U*U/2du

= 3/8∫(U^4+U^2)du

= 3/8[U^5/5+U^3/3]+C

= 3U/8[(3*U^4+5*U^2)/(3*5)]+C

= [3(4x-1)^(1/2)]/(8*15)[(4x-1)^2+(4x-1)]+C

= [(4x-1)^(1/2)][(4x-1)^2+(4x-1)]/40+C

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The indefinite integral `int 3x*(4x - 1)^(1/2) dx` has to be determined.

`int 3x*(4x - 1)^(1/2) dx`

let u = 4x - 1

x = (1 + u)/4

`dx = (1/4)*du`

=> `int (3/16)*(1 + u)*u^(1/2) du`

=> `(3/16)*int u^(1/2) + u^(3/2) du`

=> `(3/16)*u^(3/2)/(3/2) + (3/16)*u^(5/2)/(5/2) + C`

=> `(1/8)*u^(3/2) + (3/40)*u^(5/2) + C`

=> `(1/8)*(4x - 1)^(3/2) + (3/40)*(4x - 1)^(5/2) + C`

The required integral is ` ``(1/8)*(4x - 1)^(3/2) + (3/40)*(4x - 1)^(5/2) + C`