设广义积分∫（e→+无穷）f(x)dx收敛,且满足方程f(x)=2/（除以）x^2-1/（除以）x乘以lnx的平方 ∫（e→+无

设广义积分∫（e→+无穷）f(x)dx收敛,且满足方程f(x)=2/（除以）x^2-1/（除以）x乘以lnx的平方 ∫（e→+无
设广义积分∫（e→+无穷）f(x)dx收敛,且满足方程f(x)=2/（除以）x^2-1/（除以）x乘以lnx的平方 ∫（e→+无

设广义积分∫（e→+无穷）f(x)dx收敛,且满足方程f(x)=2/（除以）x^2-1/（除以）x乘以lnx的平方 ∫（e→+无
Unexpectedly only me can help you?Don't mind I say English.
Let N = ∫(e→+∞) f(x) dx,since this integral is convergent,it's a constant
f(x) = 2/x² - 1/(xln²x) · ∫(e→+∞)
f(x) = 2/x² - 1/(xln²x) · N,integrate both sides with range from e to infinity
∫(e→+∞) f(x) dx = 2∫(e→+∞) 1/x² dx - N∫(e→+∞) 1/(xln²x) dx
N = 2 · - 1/x：(e→+∞) - N∫(e→+∞) 1/ln²x d(lnx)
N = - 2 · (0 - 1/e) - N · - 1/lnx：(e→+∞)
N = 2/e + N · (0 - 1)
N = 2/e - N
2N = 2/e
N = 1/e = ∫(e→+∞) f(x) dx
So f(x) = 2/x² - 1/(xln²x) · 1/e
f(x) = 2/x² - 1/(e · xln²x)