Integration

Question

integrate ∫ 4x(x-1) / (2x-1)²

Answer 1

Let u = 2x-1  Then, x = (u+1)/2, x-1 = (u-1)/2, and dx = (1/2)du

 

Substituting into the given integral, we have 

 

   ∫4[(u+1)/2][(u-1)/2]/u²(1/2)du

 

= (1/2)∫(u²-1)/u²du

 

= (1/2)∫(1 - u^-2)du

 

= (1/2)u + (1/2)u^-1 + C₁

 

= (1/2)(2x-1) + (1/2)(2x-1)^-1 + C₁

 

= x + 1/(4x-2) + C, where C = -1/2+C₁

 

 

Answer 2

4x(x - 1) = 4x² - 4x + 1 - 1

              = (2x - 1)² - 1

 

∫ 4x(x-1) / (2x-1)² dx  = ∫ ((2x - 1)² - 1)/ (2x-1)² dx

 

                                   = ∫ (1 -(2x - 1)^-2) dx

 

                                   = x  - (-1/2)(2x - 1)^-1 + c

 

                                   = x + 1/(4x - 2) + c

 

Answer 3

The correctly written problem must look like: ∫ [4x(x-1) / (2x-1)²] dx

Given that, I= ∫ [4x(x-1) / (2x-1)²] dx

I= ∫ [(4x²-4x)/(4x²-4x=1)]dx   ............ expanding the square

 = ∫ [(4x2-4x+1-1)/(4x2-4x=1)]dx  ........... adjusting operators and coefficients

 =∫ [{(2x-1)² -1} / (2x-1)²] dx ........... adjusting operators and coefficients
 =∫dx - ∫1/ (2x-1)²] dx ........... adjusting operators and coefficients

 = x +c - I₁            ..... c is the constant of integration

Assuming 2x-1=p where p≠0

Differentiating both sides with respect to x,

2dx=dp

=>dx=dp/2 

Substituting the values,

I₁= (1/2)  ∫[1/p²] dp

  =[(1/2) (-1/p)] + c' ..... c' is the constant of integration

  =-(1/2p)+ c' 

Again substituting the values,

I=x+[1/2(2x-1)] +C  ............ C is the constant of integration