有同学经济数学二线性代数挂科了吗
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∫(3x+1)/(x^2+2x+17) dx
=(3/2)∫ (2x+2)/(x^2+2x+17) dx - 2∫ dx/(x^2+2x+17)
= (3/2)ln|x^2+2x+17| - 2∫ dx/(x^2+2x+17)
x^2+2x+17 = (x+1)^2 + 16
let
x+1= 4tany
dx = 4(secy)^2dy
∫ dx/(x^2+2x+17)
=(1/4)∫ dy
=(1/4)y+ C'
=(1/4)arctan[(x+1)/4] + C'
∫(3x+1)/(x^2+2x+17) dx
= (3/2)ln|x^2+2x+17| - 2∫ dx/(x^2+2x+17)
= (3/2)ln|x^2+2x+17| - (1/2)arctan[(x+1)/4] + C
=(3/2)∫ (2x+2)/(x^2+2x+17) dx - 2∫ dx/(x^2+2x+17)
= (3/2)ln|x^2+2x+17| - 2∫ dx/(x^2+2x+17)
x^2+2x+17 = (x+1)^2 + 16
let
x+1= 4tany
dx = 4(secy)^2dy
∫ dx/(x^2+2x+17)
=(1/4)∫ dy
=(1/4)y+ C'
=(1/4)arctan[(x+1)/4] + C'
∫(3x+1)/(x^2+2x+17) dx
= (3/2)ln|x^2+2x+17| - 2∫ dx/(x^2+2x+17)
= (3/2)ln|x^2+2x+17| - (1/2)arctan[(x+1)/4] + C
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