Question

求解 高数 (回答请标号)

Answer 1

10.
∫ arctanx / x² dx
= -∫ arctanx d(1/x)
= -(1/x)arctanx + ∫ (1/x)d(arctanx)
= -(1/x)arctanx + ∫ 1/[x(1+x²)] dx
= -(1/x)arctanx + ∫ (1/x)dx - ∫ x/(1+x²) dx
= -(1/x)arctanx + ln|x| - (1/2)ln(1+x²) + C
11.
原式=x(lnx)²-∫xd(lnx)²
=x(lnx)²-∫x*2lnx*1/xdx
=x(lnx)²-2∫lnxdx
=x(lnx)²-2xlnx+2∫xdlnx
=x(lnx)²-2xlnx+2∫x*1/xdx
=x(lnx)²-2xlnx+2∫dx
=x(lnx)²-2xlnx+2x+C
13.
令A=∫ e^(-x)sin2xdx
A=∫ e^(-x)sin2xdx = - ∫ sin2xd(e^(-x))=-e^(-x)sin2x+∫ e^(-x)d(sin2x)
=-e^(-x)sin2x+2∫ e^(-x)cos2xdx
=-e^(-x)sin2x-2∫ cos2xd(e^(-x))
=-e^(-x)sin2x-2cos2x*e^(-x)+2∫ e^(-x)d(cos2x)
=-e^(-x)sin2x-2cos2x*e^(-x)-4∫ e^(-x)sin2xdx
=-e^(-x)sin2x-2cos2x*e^(-x)-4A
5A=-e^(-x)sin2x-2cos2x*e^(-x)
A=1/5[-e^(-x)sin2x-2cos2x*e^(-x)]
=-1/5(sin2x+2cos2x)e^(-x)
14
∫ln(cosx)/cos²x dx
=∫sec²xln(cosx) dx
=∫ln(cosx)d(tanx)
=tanxln(cosx)-∫tanxd[ln(cosx)]
=tanxln(cosx)-∫tanx*1/cosx*(-sinx) dx
=tanxln(cosx)+∫tan²xdx
=tanxln(cosx)+∫(sec²x-1)dx
=tanxln(cosx)+tanx-x+C