xlnx,请问xlnx的积分怎么求？
摘要分部积分啦xlnx！过程如下：∫xlnx/[(1+x^2)^2]dx=(-1/2)∫lnxd(1/(1+x^2))=(-1/2)lnx/(1+x^2)+(1/2)∫1/[(1+x^2)*x]dx=(-
分部积分啦xlnx！
过程如下：∫xlnx/[(1+x^2)^2]dx
=(-1/2)∫lnxd(1/(1+x^2))
=(-1/2)lnx/(1+x^2)+(1/2)∫1/[(1+x^2)*x]dx
=(-1/2)lnx/(1+x^2)+(1/2)∫x/[(1+x^2)*x^2]dx
=(-1/2)lnx/(1+x^2)+(1/4)∫1/[(1+x^2)*x^2]d(x^2)
=(-1/2)lnx/(1+x^2)+(1/4)∫[1/x^2-1/(1+x^2)]d(x^2)
=(-1/2)lnx/(1+x^2)+(1/4)[ln(x^2)-ln(1+x^2)]+C
=(-1/2)lnx/(1+x^2)+(1/4)ln[x^2/(1+x^2)]+C
^1.在【1.3】上f'(x)=lnx+1>0,f(x)单调递增，最小值为f(1)=0
2.a≤2lnx+x+3/x,令g(x)=2lnx+x+3/x,x∈[1/e,e]
g'(x)=2/x+1-3/x^2=(x+3)(x-1)/x^2,故在[1/e,1]上，g'(x)<0,g(x)单调递减，在[1，e]上，g'(x)>0,g(x)单调递增，g(e)=2+e+3/e<g(1/e)=3e+1/e-2,在【1/e，e】g(x)的最大值为3e+1/e-2
所以a≤3e+1/e-2
