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Evaluate the integral `int (x*e^(2x))/(2x+1)^2 dx` .

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user8052413 | eNoter

Posted January 30, 2013 at 2:45 AM via web

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Evaluate the integral `int (x*e^(2x))/(2x+1)^2 dx` .

Tagged with calculus, integration

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 28, 2013 at 1:32 PM (Answer #1)

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The integral `int (x*e^(2x))/(2x+1)^2 dx` has to be determined. Use integration by parts.

Let `x*e^(2x) = u`

`du = 2*x*e^(2x) + e^(2x) dx`

=>` du = e^(2x)*(1 + 2x) dx`

`dx/(2x+1)^2 = dv`

=> `dv = ((1/2)*2)/(2x+1)^2 dx`

=> `dv = (1/2)*(d(2x + 1))/(2x + 1)^2`

=> `dv = (1/2)(2x + 1)^(-2)*d(2x + 1) `

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

=> `v = -1/(2*(2x+1))`

  `int (x*e^(2x))/(2x+1)^2 dx`

= `u*v - int v du`

= `x*e^(2x)*((-1)/(2*(2x+1))) - int (-1)/(2*(2x+1)) du`

= `x*e^(2x)*((-1)/(2*(2x+1))) - int (-1)/(2*(2x+1))*e^(2x)*(2x+1) dx`

= `x*e^(2x)*((-1)/(2*(2x+1))) + int e^(2x)/2 dx`

= `x*e^(2x)*((-1)/(2*(2x+1))) + (e^(2x))/4 + C`

=> `(e^(2x)/2)((1/2) - x/(2x+1)) + C`

=> `(e^(2x)/2)((2x + 1 - 2x)/(2*(2x+1))) + C`

=> `e^(2x)/(4*(2x+1)) + C`

The required integral is `e^(2x)/(4*(2x+1)) + C`

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