Question

这几个题，求解

Answer 1

∫xe^(2x)dx
=(1/2)∫xd(e^2x)
=(1/2)[x·e^(2x)-∫e^(2x)dx]
=(1/2)x·e^(2x)-(1/2)·(1/2)∫e^(2x)d(2x)
=(1/2)x·e^(2x)-(1/4)e^(2x)+C
∫xcos3xdx=(1/3)∫x·cos3xd(3x)
=(1/3)∫xd(sin3x)
=(1/3)[x·sin3x-∫sin3xdx]
=(1/3)x·sin3x+(1/9)cos3x+C
∫xarctanxdx
=(1/2)∫arctanxd(x²)
=(1/2)[x²·arctanx-∫x²d(arctanx)]
=(1/2)x²·arctanx-(1/2)∫[x²/(1+x²)]dx
=(1/2)x²·arctanx-(1/2)∫[(x²+1-1)/(1+x²)]dx
=(1/2)x²·arctanx-(1/2)x+(1/2)arctanx+C
=(1/2)[x²·arctanx-x+arctanx]+C
令∫e^(-x)cos2xdx=A
则A=-∫cos2xd(e^-x)
=-[e^(-x)·cos2x-∫e^(-x)d(cos2x)]
=-[e^(-x)·cos2x+2∫e^(-x)·sin2xdx]
=-e^(-x)·cos2x-2∫e^(-x)·sin2xdx
=-e^(-x)·cos2x+2∫sin2xd(e^-x)
=-e^(-x)·cos2x+2[e^(-x)·sin2x-∫e^(-x)d(sin2x)]
=-e^(-x)·cos2x+2e^(-x)·sin2x-4∫e^(-x)·cos2xdx
=-e^(-x)·(cos2x-2sin2x)-4A
所以，5A=-e^(-x)·(cos2x-2sin2x)+C1
即，原式=A=(-1/5)·e^(-x)·(cos2x-2sin2x)+C
令(1+x)^(1/6)=t，则1+x=t^6，dx=6t^5tdt
且√(1+x)=t³，(1+x)^(1/3)=t²
原式=∫[t³/(1+t²)]·6t^5dt
=6∫[t^8/(1+t²)]dt
=6∫[(t^8-1)+1]/(1+t²)dt
=6∫[(t^4+1)·(t²+1)·(t²-1)+1]/(1+t²)dt
=6∫(t^4+1)·(t²-1)dt+6∫[1/(1+t²)]dt
=6∫(t^6-t^4+t²-1)dt+6arctant
=(6/7)t^7-(6/5)t^5+2t³-6t+6arctant+C——【代入t即可】

Answer 2

4. 要查看其中的密码，必须要求手机具有root权限