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求不定积分E根号x减1

令其等于t得t方加一分之2t方的不定积分,二提出t方加一减一,再反代换

设根号(e^x-1) =t t^2 +1=e^x x=ln(t^2 +1) 代入得 ∫t dln(t^2 +1) =∫2t^2/(t^2 +1) dt =2*∫t^2/(t^2 +1) dt =2*∫(t^2 +1-1)/(t^2 +1) dt =2*∫[1 -1/(t^2 +1)] dt =2*[∫1 dt -∫1/(t^2 +1) dt =2*(t -arctant) +C(常数) =2*【(e^x-1) -arctan(e^x-1)】+C=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)

积分符号 我用f代替了 令t=根号(1+e^x) 那么x=in(t^2-1) 所以dx=din(t^2-1)=2t/(t^2-1)dt 那么原积分可以写成f2dt/(t^2-1)=2f1/t^2-1 dt=ln{[根号(1+e^x)-1]/[根号(1+e^x)+1]}+c

令√(e^x-1)=t e^x-1=t^2 e^x=t^2+1 x=ln(t^2+1) dx=2t/(t^2+1) x=0,则t=0 x=ln2,则t=1 原式=∫(0,1)2t^2dt/(t^2+1) =2∫(t^2+1-1)dt/(t^2+1) =2∫[1-1/(t^2+1)]dt =2t-2ln(t^2+1) =2-0-2ln2+2ln1 =2-2ln2

换元:t=根号(ex-1)就行了得出:=2(x-arctanex)+c

t=(e^x+1)^0.5 dx=2t/(t^2-1) ∫(e^x+1)^0.5 dx=∫2t^2/(t^2-1)dt=∫2 +2/(t^2-1)dt=2t+ln[(t-1)/(t+1)]+c

解: 设根号(e^x-1) =t t^2 +1=e^x x=ln(t^2 +1) 代入得 ∫t dln(t^2 +1) =∫2t^2/(t^2 +1) dt =2*∫t^2/(t^2 +1) dt =2*∫(t^2 +1-1)/(t^2 +1) dt =2*∫[1 -1/(t^2 +1)] dt =2*[∫1 dt -∫1/(t^2 +1) dt =2*(t -arctant) +C(常数) =2*【(e^x-1) -arctan(e^x-1)】+C=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)

∫e^√x dx 令√x=t x=t^2 dx=2tdt 原式=∫e^t*2tdt=2∫tde^t=2te^t-2∫e^tdt=2te^t-2e^t+c=2e^t(t-1)+c=2e^√x(√x-1)+c

答:设t=√(x-1),x=t^2+1,dx=2tdt ∫ e^[√(x-1)] dx=∫ e^t *2t dt=2te^t-2∫e^t dt=2te^t-2e^t+C=2√(x-1)*e^[√(x-1)-2e^[√(x-1)]+C

设t=e^根号(x+1) 则x=(lnt)^2-1 dx=(2lntdt)/t∫(e^根号(x+1)) dx=∫t*(2lntdt)/t=∫2lntdt=2∫lntdt=2tlnt-t+C=2e^根号(x+1)*根号(x+1)-e^根号(x+1)+C

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