Question

请帮忙算一下下面的不定积分

1、∫xe^xcosxdx
2、∫[（xe^x）/√(1+e^x)]dx
3、∫[（x+sinx）/(1+cosx)]dx

请给出详细步骤
先谢谢了!！

Answer 1

1.使用分部积分法，得出循环后移项

∫x(e^x)cosxdx
=∫x(e^x)dsinx
=x(e^x)sinx-∫sinxd[x(e^x)]
=x(e^x)sinx-∫[e^(x)+xe^(x)]sinxdx
=x(e^x)sinx-∫e^(x)sinxdx-∫xe^(x)sinxdx
=x(e^x)sinx-∫sinxde^(x)+∫xe^(x)dcosx
=x(e^x)sinx-e^(x)sinx+∫e^(x)dsinx+xe^(x)cosx-∫cosxd[xe^(x)]
=x(e^x)sinx-e^(x)sinx+∫e^(x)cosxdx+xe^(x)cosx-∫[e^(x)+xe^(x)]cosxdx
=x(e^x)sinx-e^(x)sinx+∫e^(x)cosxdx+xe^(x)cosx-∫e^(x)cosxdx-∫xe^(x)cosxdx
=x(e^x)sinx-e^(x)sinx+xe^(x)cosx-∫xe^(x)cosxdx

∫xe^(x)cosxdx=[x(e^x)sinx-e^(x)sinx+xe^(x)cosx]/2+C

2.第一换元法(凑微分法)、第二换元法、分部积分法一起使用

首先求不定积分∫1/√(1+e^x)dx
令√(1+e^x)=t，则x=ln(t²-1)=ln(t+1)(t-1)=ln(t+1)+ln(t-1)，dx=[1/(t+1)+1/(t-1)]dt
∫1/√(1+e^x)dx
=∫1/t(t+1)dt+∫1/t(t-1)dt
=∫1/[(t+1/2)²-(1/2)²]dt+∫1/[(t-1/2)²-(1/2)²]dt
=∫1/[(t+1/2)²-(1/2)²]d(t+1/2)+∫1/[(t-1/2)²-(1/2)²]d(t-1/2)
=ln|(t+1/2-1/2)/(t+1/2+1/2)|+ln|(t-1/2-1/2)/(t-1/2+1/2)|
=ln|t/(t+1)|+ln|(t-1)/t|
=ln|(t-1)/(t+1)|
=ln[√(1+e^x)-1]/[√(1+e^x)+1]

∫e^x/√(1+e^x)dx
=∫1/√(1+e^x)de^x
=2√(1+e^x)

∫xe^x/√(1+e^x)dx
=2∫xd√(1+e^x)
=2x√(1+e^x)-2∫√(1+e^x)dx
=2x√(1+e^x)-2∫(1+e^x)/√(1+e^x)dx
=2x√(1+e^x)-2∫1/√(1+e^x)dx-2∫e^x/√(1+e^x)dx
=2x√(1+e^x)-2ln[√(1+e^x)-1]/[√(1+e^x)+1]-2√(1+e^x)+C

3.第一换元法(凑微分法)、分部积分法一起使用

∫(x+sinx)/(1+cosx)dx
=∫[x+2sin(x/2)cos(x/2)]/2cos²(x/2)dx
=∫x/2cos²(x/2)dx+∫tan(x/2)dx
=∫xsec²(x/2)d(x/2)+∫tan(x/2)dx
=∫xdtan(x/2)+∫tan(x/2)dx
=xtan(x/2)-∫tan(x/2)dx+∫tan(x/2)dx
=xtan(x/2)+C