Question

求这四道积分题的详细解答过程

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Answer 1

∫sin^2xcos^2xdx
=(1/4)*∫sin^2(2x)dx
=(1/8)*∫(1-cos4x)dx
=(1/8)*[x-(1/4)*sin4x]+C
=x/8-(1/32)*sin4x+C，其中C是任意常数
∫xe^(-x)dx
=-∫xd[e^(-x)]
=-xe^(-x)+∫e^(-x)dx
=-xe^(-x)-e^(-x)+C
=-(x+1)e^(-x)，其中C是任意常数
∫x^4/(x^2-1)dx

=∫[(x^4-1)/(x^2-1)+1/(x^2-1)]dx
=∫[x^2+1+1/(x+1)(x-1)]dx
=∫[x^2+1+1/2(x-1)-1/2(x+1)]dx
=(1/3)*x^3+x+(1/2)*ln|x-1|-(1/2)*ln|x+1|+C，其中C是任意常数
∫1/x(1+x^5)dx
令t=1/x，则x=1/t，dx=-dt/t^2
=∫t/(1+1/t^5)*(-dt/t^2)
=-∫t^4/(t^5+1)dt
=(-1/5)*∫d(t^5+1)/(t^5+1)
=(-1/5)*ln|t^5+1|+C
=(-1/5)*ln|1/x^5+1|+C，其中C是任意常数