Question

求1为上限,-1为下限的定积分∫e^x/(e^x+1)dx

Answer 1

∫e^x/(1+e^2x)^2 dx=∫(e^x+e^3x-e^3x)/(1+e^2x)^2 dx=∫e^x/(1+e^2x)-e^3x/(1+e^2x)^2 dx=∫e^x/(1+e^2x)dx-∫e^3x/(1+e^2x)^2 dx=∫1/(1+e^2x)de^x-∫e^2x/(1+e^2x)^2 de^x=tane^x-∫e^2x/(1+e^2x)^2 de^x
令t=e^x,u=t+1
∫e^2x/(1+e^2x)^2 de^x=∫t^2/(1+t)^2dt=∫(t^2-1+1)/(1+t)^2dt=∫(t-1)/(t+1)+1/(t+1)^2dt=∫1-2/(t+1)+1/(t+1)^2d(t+1)=u-2ln|u|-1/u+c=e^x+1-2ln|e^x+1|-1/(e^x+1)+c
原式=lim∞>∫e^x/(1+e^2x)^2 dx=lim∞>tane^b-tane^a-e^b+e^a+2ln|(e^b+1)/(e^a+1)|+1/(e^b+1)-1/(e^a+1)=-tane^a+e^a+∞-1/(e^a+1)=∞

Answer 2

∫[-1,1]e^x/(e^x+1)dx

=∫[-1,1]1/(e^x+1)de^x
=ln(e^x+1)[-1,1]
=ln(e+1)-ln(1/e+1)

追问

∫1/(3x-2)^2 dx,(下限为0,上限为2)

追答

∫[0,2]1/(3x-2)^2 dx

=-1/3*1/(3x-2)[0,2]
=-1/4

追问

∫1/(x^2-5x-6) dx,(下限为负无穷,上限为正无穷)