2道不定积分的题∫(e^x-1)^1/2dx∫coslnxdx

1个回答

  • 1.∫(e^x-1)^1/2dx

    设(e^x-1)^1/2=t

    t^2=e^x-1,x=ln(t^2+1)

    ∫(e^x-1)^1/2dx =∫tdln(t^2+1)

    =∫2t^2/(t^2+1)dt

    =∫(2t^2+2-2)/(t^2+1)dt

    =2-2∫1/(t^2+1)dt+C

    =2-2arctan(t) +C

    =2-2arctan[(e^x-1)^1/2] +C

    2.

    ∫coslnxdx=(分部积分) xcoslnx-∫xdcoslnx

    =xcoslnx+∫xsinlnx/x dx

    =xcoslnx+∫sinlnx dx

    =(分部积分) xcoslnx+xsinlnx-∫xdsinlnx

    =xcoslnx+xsinlnx-∫coslnx dx +C

    所以 2∫coslnx dx=xcoslnx+xsinlnx +C

    ∫coslnx dx=x(coslnx+sinlnx)/2 +C