Solve the indefinite integral (2x/(3-x²))
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Solve the indefinite integral (2x/(3-x²))

[From: ] [author: ] [Date: 12-02-21] [Hit: ]
du = -2x dx, or dx = du/-2x. you are left with the integral of (2x/u) * (du/-2x). you are left with the integral of -du/u, which is -ln(u) + C. substitute the u back in and you get -ln(3-x^2) + C.......
use u substitution. set u = 3 - x^2. Therefore, du = -2x dx, or dx = du/-2x. you are left with the integral of (2x/u) * (du/-2x). you are left with the integral of -du/u, which is -ln(u) + C. substitute the u back in and you get -ln(3-x^2) + C.

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∫ 2x/(3-x^2) dx

Let u=3-x^2
du =-2x dx
2x dx = -du

∫ 2x/(3-x^2) dx = - ∫ du/u = -ln(u)+C = - ln(3-2x^2) + C
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keywords: sup,indefinite,the,integral,Solve,Solve the indefinite integral (2x/(3-x²))
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