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∫e∧x-1/e∧x+1dx的不定积分

∫(e^x-1)/(e^x+1) dx =∫1 -2/(e^x+1) dx =∫1 -2/(e^x+1)e^x de^x =x -2ln[e^x/(e^x+1)] +C,C为常数

∫ex?1ex+1dx=∫exex+1dx?∫1ex+1dx=∫exex+1dx?∫e?x1+e?xdx=∫1ex+1d(ex+1)+∫11+e?xd(1+e?x)=ln(1+ex)+ln(1+e-x)+C.

-1/2e^-2x+x

1、令[根号(x+1)]=t,则x=t^2-1,dx=2tdt,所以 原式=∫(t^2-1)t×2tdt =∫(2t^4-2t^2)dt =(2/5)t^5-(2/3)t^3+C =(2/5)[(x+1)^(5/2)]-(2/3)[(x+1)^(3/2)]+C 2、∫lnxdx =xlnx-∫xd(lnx) =xlnx-x+C 所以, ∫lnxdx =(xlnx-x+C)| =1 3、令F(x,y,z)=(x^...

标题没看懂,但是你的图可以看懂

解:原式=积分(e^x+1)(e^x-1)/(e^x+1)dx =积分e^x-1dx =积分e^xdx-积分1dx =e^x-x+C 答:原函数为e^x-x+C。

=∫(x+1/2)/(x²+x+1)dx+1/2∫1/((x+1/2)²+3/4)dx =1/2∫1/(x²+x+1)d(x²+x+1)+1/2∫1/(u²+3/4)du =(1/2)ln(x²+x+1)+(1/2)/(3/4)*√3/2*arctan(2u/√3)+C =(1/2)ln(x²+x+1)+(1/√3)arctan((2x+1)/√3)+C

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