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
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