can you integrate this....?

very easy girl...

see ,

first use, 4x-1=t^2;

then ,

integration willl become-

(3/4) * t (t^2 + 1);

then open all the brackets and solve;

i think now u can do it.

3x( 4x - 1)^1/2

let ( 4x - 1)^1/2 = t

4x - 1 = t^2

x = 1/4( t^2 + 1 ) . . . . . substitute

3x( 4x - 1)^1/2 = 3/4 ( t^2 + 1 ) t^2

integ 3/4 ( t^4 + t^2 ) dt = 3/4 ( t^5/5 + t^3/3 )

. . . . = 3/4 [ (4x -1)^3/2 + 4x -1 ]

fine..

let 4x-1=t^2.

it implies 4dx= 2tdt implies dx=t/2 *dt.

now the integrand becomes 3*[t^2+1]/4 *t * t/2*dt

now u can proceed ..........( just open the brackets and integration becomes easy.

actually its a difficult job to write down each step.

so a little hint may be sufficient for catching the problem.

in(3x(4x-1)^1/2)dx

Let u^2=4x-1

2udu=4dx

4x=u^2-1

3/8*int( (u^2+1)*u*u) du

=3/8 in(u^4+u^2 )du = 3/8*(u^5/5+u^3/u)+c

where u=sqrt()4x-1)

lets take 4x-1=t^2

x= t^2/4+1/4

dx= 2t/4=t/2

now 3x(4x-1)^1/2 becomes 3(t^2/4+1/4)(t)(t/2)

expanding 3/8t^4+3/8t^2

integration gives you 3/40 (t^5) + 3/24 (t^3)

ie 3/4(t^3)(t^2+1/6)

ie 3/4(4x-1)^3/2(4x-5/6)

are there limits? or is it purely S 3x(4x-1)^1/2 dx