Given that if y=x^2/(2x-1)^2,dy/dx=-2x/(2x-1)^3, find the integral of 4x-3/(2x-1)^3

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The integral of -2x/(2x - 1)^3 is x^2/(2x - 1)^2, we have to find the integral of 4x-3/(2x-1)^3

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

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

=> -2*(-2x/(2x - 1)^3 - 3/(2x - 1)^3

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

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

let 2x - 1 = t => dt = 2*dx => 3*dx = (3/2) dt

=> `(-2x^2)/(2x - 1)^2 - (3/2)*int 1/(t^3) dt`

=> `(-2x^2)/(2x - 1)^2 - (3/2)/(-2*t^2)`

substitute t = 2x - 1

=> `(-2x^2)/(2x - 1)^2 + 3/(4*(2x - 1)^2)`

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

 

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