How do you integrate this
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How do you integrate this

[From: ] [author: ] [Date: 11-05-14] [Hit: ]
dv = e^(2x) dx ==> v = (1/2)e^(2x).= e^(2x)/[4(2x + 1)] + C.(Note that the two 1/2 ∫ e^(2x)/(2x + 1)^2 dxs cancel out.I hope this helps!-Youre welcome! :)Report Abuse -int(x*exp(2*x)/(2*x+1)^2,......
integral (xe^(2x) / (2x+1)^2)

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To integrate ∫ x*e^(2x)/(2x + 1)^2 dx, note that:
∫ x*e^(2x)/(2x + 1)^2 dx
= 1/2 ∫ 2x*e^(2x)/(2x + 1)^2 dx
= 1/2 ∫ [(2x + 1) - 1]e^(2x)/(2x + 1)^2 dx
= 1/2 ∫ [1/(2x + 1) - 1/(2x + 1)^2]e^(2x) dx
= 1/2 ∫ [e^(2x)/(2x + 1) - e^(2x)/(2x + 1)^2] dx
= 1/2 ∫ e^(2x)/(2x + 1) dx - 1/2 ∫ e^(2x)/(2x + 1)^2 dx.

Then, integrate ∫ e^(2x)/(2x + 1) dx by parts with:
u = 1/(2x + 1) ==> du = -2/(2x + 1)^2 dx
dv = e^(2x) dx ==> v = (1/2)e^(2x).

This gives:
∫ e^(2x)/(2x + 1) dx = uv - ∫ v du
= e^(2x)/[2(2x + 1)] - ∫ -e^(2x)/(2x + 1)^2 dx
= e^(2x)/[2(2x + 1)] + ∫ e^(2x)/(2x + 1)^2 dx

Therefore:
∫ x*e^(2x)/(2x + 1)^2 dx
= 1/2 ∫ e^(2x)/(2x + 1) dx - 1/2 ∫ e^(2x)/(2x + 1)^2 dx
= e^(2x)/[4(2x + 1)] + 1/2 ∫ e^(2x)/(2x + 1)^2 dx - 1/2 ∫ e^(2x)/(2x + 1)^2 dx
= e^(2x)/[4(2x + 1)] + C.
(Note that the two 1/2 ∫ e^(2x)/(2x + 1)^2 dx's cancel out.)

I hope this helps!

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You're welcome! :)

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int(x*exp(2*x)/(2*x+1)^2, x) = (1/4)*exp(2*x)/(2*x + 1)
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