f(x) ＝log（1/x）x>0 求 ∫xf(x)dx

f(x) ＝log（1/x）x>0 求 ∫xf(x)dx
f(x) ＝log（1/x）x>0 求 ∫xf(x)dx

f(x) ＝log（1/x）x>0 求 ∫xf(x)dx
假如㏒(1/x)是以底为10,真数为1/x的对数
则∫xf(x)dx
=∫x㏒(1/x)dx
=∫xln(1/x)/ln10 dx,换底公式
=(1/ln10)∫xln(x^-1)dx
=(-1/ln10)∫xlnxdx
=(-1/ln10)(1/2)∫lnxd(x²),分部积分
=(-1/2ln10)[x²lnx-∫x²d(lnx)]
=(-1/2ln10)(x²lnx-∫xdx)
=(-1/2ln10)(x²lnx-x²/2)+C
=x²/(4ln10)-x²lnx/(2ln10)+C
亦可以转换为x²/(4ln10)-x²㏒x/2+C

令1/x=t 则x=1/t f(x)=log(t)(1/t) dx=d(1/t)
∫xf(x)dx=∫1/tlog(t)(1/t)d(1/t)
=1/2∫log(t)(1/t)d(1/t^2)
=1/2{log(t)(1/t)(1/t^2)+∫(1/t)[1/(1/t)lnt]d(1/t)}
=1/2[log(t)(1/t)(1/t^2)+∫(t/lnt)d...

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令1/x=t 则x=1/t f(x)=log(t)(1/t) dx=d(1/t)
∫xf(x)dx=∫1/tlog(t)(1/t)d(1/t)
=1/2∫log(t)(1/t)d(1/t^2)
=1/2{log(t)(1/t)(1/t^2)+∫(1/t)[1/(1/t)lnt]d(1/t)}
=1/2[log(t)(1/t)(1/t^2)+∫(t/lnt)d(1/t)
=1/2[log(t)(1/t^3)+∫(t/lnt)d(1/t)
=1/2[log(t)t^(-3)-∫(t/lntt^2)dt]
=1/2(-3)-∫(1/tlnt)dt
=-3/2-∫(1/lnt)d(lnt)
=-3/2-∫d(lnt)=-3/2-lnt=-3/2+lnx+C

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