五个不定积分题。求一下答案？

Question

如题所述

Answer 1

（1）令x=tant，则dx=sec^2tdt
原式=∫ln(tant+sect)/sect*sec^2tdt
=∫ln(tant+sect)*sectdt
=∫ln(tant+sect)d[ln(tant+sect)]
=(1/2)*[ln(tant+sect)]^2+C
=(1/2)*[ln(x+√(1+x^2)]^2+C，其中C是任意常数
（2）原式=∫e^(sin2x-2x)*sin^2xdx
=(1/2)*∫e^(sin2x-2x)*(1-cos2x)dx
=(1/2)*∫e^(sin2x-2x)d[x-(1/2)*sin2x]
=(-1/4)*∫e^(sin2x-2x)d(sin2x-2x)
=(-1/4)*e^(sin2x-2x)+C，其中C是任意常数
（3）原式=∫[(1-x)/(x-lnx)^2+1/(x-lnx)]dx
=∫(1-x)/(x-lnx)^2dx+∫dx/(x-lnx)
=∫(1-x)/(x-lnx)^2dx+x/(x-lnx)-∫xd[1/(x-lnx)]
=∫(1-x)/(x-lnx)^2dx+x/(x-lnx)+∫x(1-1/x)/(x-lnx)^2
=∫(1-x)/(x-lnx)^2dx+x/(x-lnx)+∫(x-1)/(x-lnx)^2
=x/(x-lnx)+C，其中C是任意常数
（4）原式=∫e^x*[sin(x/2)+cos(x/2)]^2/[2cos^2(x/2)]dx
=(1/2)*∫e^x*[tan(x/2)+1]^2dx
=(1/2)*∫e^x*[tan^2(x/2)+2tan(x/2)+1]dx
=(1/2)*∫e^x*sec^2(x/2)dx+∫e^x*tan(x/2)dx
=∫e^x*d[tan(x/2)]+∫e^x*tan(x/2)dx
=e^x*tan(x/2)-∫e^x*tan(x/2)dx+∫e^x*tan(x/2)dx
=e^x*tan(x/2)+C，其中C是任意常数
（5）令t=x^3，则dt=3x^2dx
原式=(1/3)*∫(x^3*3x^2dx)/√(a^3-x^3)
=(1/3)*∫tdt/√(a^3-t)
令u=√(a^3-t)，则t=a^3-u^2，dt=-2udu
原式=(1/3)*∫[(a^3-u^2)/u]*(-2u)du
=(2/3)*∫(u^2-a^3)du
=(2/3)*[(1/3)*u^3-a^3*u]+C
=(2/9)*(a^3-t)^(3/2)-(2/3)*a^3*√(a^3-t)+C
=(2/9)*(a^3-x^3)^(3/2)-(2/3)*a^3*√(a^3-x^3)+C
=(-2/9)*√(a^3-x^3)*(2a^3+x^3)+C，其中C是任意常数