嘿嘿,其實這題很簡單.
令y = 1/x、x = 1/y、dx = - 1/y² dy
∫ [arctan(1/x)]/(1 + x²) dx
= ∫ arctany/(1 + 1/y²) * (- 1/y² dy)
= ∫ arctany * y²/(1 + y²) * (- 1/y²) dy
= - ∫ arctany/(1 + y²) dy
= - ∫ arctany d(arctany)
= (- 1/2)(arctany)² + C
= (- 1/2)[arctan(1/x)]² + C為啥我就想不到- -可能是你練習太少吧。這種題目每天應(yīng)該做100題,做熟就會了。100道??沒夸張???額,可能對你不適合,這是讀數(shù)學(xué)專業(yè)的要求但是你還得多練習才是。起碼要看熟題型用哪種方法,以后做題快很多了謝謝謝謝~~~~~~~~~~~~~~~
求積分∫(arctan(1/x)/(1+x^2))dx
求積分∫(arctan(1/x)/(1+x^2))dx
數(shù)學(xué)人氣:504 ℃時間:2020-02-02 18:05:20
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