※ 引述《a224996 (綠茶)》之銘言：
: 1. ∫xlnx dx
let u = lnx , dv = x dx
du = dx/x , v = x^2/2
∫xlnx dx
= (x^2/2)lnx - ∫(x/2) dx
= (x^2/2)lnx - (x^2/4) + C
: 2. ∫x/1+x^2 dx 上下極限 正負無窮大
: 請幫解...
let u = 1+x^2 , du = 2x dx
∫x/1+x^2 dx
= ∫(1/u)(1/2)du
= (1/2)∫du/u
= (1/2)lnu + C
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