Question

integration of (2x - 1 )/ (4x^2 +4x + 2)^1/2

Answer 1

Hey there !!!!

$\int\ (2x-1)/ (4x^2+4x+2) ^{1/2} \, dx$------Equation 1

let ,
        4x²+4x+2=t²

            differentiating wrt to t
    
        (8x+4)dx=2tdt

                  dx= tdt/2(2x+1)

Substituting  dx= tdt/2(2x+1) and 4x²+4x+2=t² in equation 1

$\int\ (2x-1)t/2(2x+1)t \, dt$

$\int\ (2x-1)/(2x+1) \, dt$

Now splitting the numerator

$\int\ ((2x/2x+1)-(1/2x+1) )\, dx$

Now adding and subtracting 1 in the first part of numerator

$\int\ ( 2x-1+1/2x+1)-(1/2x+1) \, dt$



Further splitting the first part 

$\int\ ((2x+1/2x+1)-(1/2x+1)-(1/2x+1 ))\, dt$

$\int\ ( 1-(2/2x+1)) \, dt$ ----- Equation 2

But 4x²+4x+2=t²

   (2x+1)²=t²-1

 (2x+1)=√t²-1

Now substituting the above in equation 2

$\int\ ( 1-(2/2x+1)) \, dt$

$\int\ (1-2/ \sqrt{t^2-1} \, dt$ 

=t-2cosh⁻¹ t

 =$\sqrt{4x^{2}+ 4x+2}$-2cosh⁻¹($\sqrt{4x^{2}+ 4x+2}$)

Hope this helped you ......................



                

        

Answer 2

I = (2x -1)dx/√(4x² + 4x + 2)

(4x² + 4x + 2) = P { let }
differentiate wrt x
8x + 4 = dP/dx
(8x + 4)dx = dP

now,

I ={ 2/8.(8x + 4) -2 }dx /√(4x² + 4x +2)

=1/4.(8x + 4)dx/√(4x² + 4x +2) -2dx/√(4x² + 4x +2)

put above results

I =1/4. dp/√P - 2dx/√(4x² + 4x +2)

I = 1/2.√P - 2dx/√{(2x + 1)² +1² }

I = 1/2√(4x² +4x + 2) -2dx/√{(2x +1)² +1²}

[ use, dx/√(x² + a² ) = ln(x + √(x² + a²) ) ]

I = 1/2.√(4x² + 4x +2) - 2/2ln{( 2x + 1 + √(4x² + 4x +2) }

I = 1/2.√(4x² +4x +2) -ln{ 2x +1 +√(4x²+4x +1) } + C