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5x 若是乘以分式,则乘在分子上,
I = ∫5xdx/(x^2+4) = (5/2)∫d(x^2+4)/(x^2+4) = (5/2)ln(x^2+4) + C.
5x 若是乘在分母上,
1/[5x(x^2+4)] = (1/5)(1/4)[1/x - x/(x^2+4)] = (1/20)[1/x - x/(x^2+4)]
I = ∫dx/[5x(x^2+4)] = (1/20)∫[1/x - x/(x^2+4)]dx
= (1/20)[ln|x| - (1/2)ln(x^2+4)] + C = (1/40)ln[x^2/(x^2+4)] + C
I = ∫5xdx/(x^2+4) = (5/2)∫d(x^2+4)/(x^2+4) = (5/2)ln(x^2+4) + C.
5x 若是乘在分母上,
1/[5x(x^2+4)] = (1/5)(1/4)[1/x - x/(x^2+4)] = (1/20)[1/x - x/(x^2+4)]
I = ∫dx/[5x(x^2+4)] = (1/20)∫[1/x - x/(x^2+4)]dx
= (1/20)[ln|x| - (1/2)ln(x^2+4)] + C = (1/40)ln[x^2/(x^2+4)] + C
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