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whats the answer of this integral: ∫ 5x/√(3-2x) dx ??

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  • whats the answer of this integral: ∫ 5x/√(3-2x) dx ??


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Answer #1 | 23/12 2013 07:22
LOL...not something I'd want to try scribbling out...that's why they make integral tables. :)
Answer #2 | 23/12 2013 07:40
∫ 5x/√(3-2x) dx Let t = 3-2x dt = - 2 dx dx = (-1/2) dt -2x = t-3 2x=3-t x=(1/2)(3-t) ∫ 5x/√(3-2x) dx = (-5/2)(1/2) ∫ (3-t) / t^(1/2) dt = (-5/4) ∫ (3t^(-1/2) -t^(1/2)) dt = (-15/4) ∫ t^(-1/2) dt + (5/4) ∫ t^(1/2) dt = (-15/4) t^(-1/2+1)/(-1/2+1) + (5/4) t^(1/2+1)/(1/2+1) = (-15/4)(2)t^(1/2) + (5/4)(2/3) t^(3/2) = (-15/2) t^(1/2) +(5/6) t^(3/2) replace t by 3-2x = (-15/2) (3-2x)^(1/2) + (5/6) (3-2x)^(3/2) = (-15/2) sqrt(3-2x) + (5/6) (3-2x)^(3/2) + C
Answer #3 | 23/12 2013 09:07
5x*(3-2x)^(-1/2)=t(x) try z(x)=3-2x^1/2 derivative is 3-2x^-1/2*-1=z'(x) z'(x)=t(x)*-1/5x and z(x)=z'(x)*(2x-3) try xz(x) derivative is xz'(x)+z(x)=z'(x)*(3x-3) you need to get z'(x)*-5x so derivative of (ax+b)z(x) is z'(x)*(b+3ax-3a) a=-5/3 b=-5 answer is z(x)(-5/3x-5)

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