※ 引述《a224996 (綠茶)》之銘言：
: 1. ∫xlnx dx
: 2. ∫x/1+x^2 dx 上下極限 正負無窮大
: 請幫解...
1. Let u = lnx , dv = x dx
Using integration by parts, we get ∫u dv = uv - ∫v du
= (x^2/2)(ln x) - ∫(x^2/2)(1/x)dx
= (x^2/2)(ln x) - ∫(x/2)dx
= (x^2/2)(ln x) - x^2/4
2. Consider the improper integral with lower limit 0
upper limit ∞ firstly.
Note that Σ(x/1+x^2) diverges since Σ(1/x) diverges and
(1/x)/(x/(1+x^2))→1 as x→∞.(Limit Comparison Test)
Thus ∫x/1+x^2 dx with lower limit 0 upper limit ∞ diverges
and hence the required improper integral diverges.
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