已知f(x)的原函数为(lnx)^2,求∫ xf'(x)dx

ƒ(x)的原函数为(lnx)²
==> ∫ ƒ(x) dx = (lnx)²
==> ƒ(x) = 2(lnx)(1/x) = (2/x)(lnx)
∫ xƒ'(x) dx
= ∫ x d[ƒ(x)]
= xƒ(x) - ∫ ƒ(x) dx
= x(2/x)(lnx) - (lnx)²
= 2lnx - (lnx)²∫ ƒ(x) dx = (lnx)²==> ƒ(x) = 2(lnx)(1/x) = (2/x)(lnx)这步是什么，看不懂哦？为什么就可以推出f(x)？？两边求导[∫ ƒ(x) dx]' = ƒ(x)[(lnx)²]' = d(lnx)²/d(lnx) • d(lnx)/dx = 2lnx • 1/x没事了，谢谢，懂了嗯~