有關微積分一些題目~真的很危機不太會(微積分高手請近)

解到一半不會解 >.< 再接再厲, 不達目的, 誓不罷休

2008-04-30 01:57:10 補充：
∫ ( x^(1/2) + e^√x )^2 dx

= ∫ x dx + 2 ∫ x^(1/2)e^√x dx + ∫ e^(2√x) dx

由於太複雜, 我就分三段來積分, 最後再合併

∫ x dx = x^2/2 + C1

2 ∫ x^(1/2)e^√x dx

令 u = 2x, dv = e^√x/(2x^(1/2)) dx

則 du = 2 dx, v = e^√x

2 ∫ x^(1/2)e^√x dx

= 2 * ( 2xe^√x – ∫ 2 e^√x dx )

= 2 ( 2xe^√x – 2 * ∫ 2x^(1/2)e^√x/(2x^(1/2)) dx )

令 u = 2x^(1/2), dv = e^√x/(2x^(1/2)) dx

則 du = 1/x^(1/2) dx, v = e^√x

∫ 2x^(1/2)e^√x/(2x^(1/2)) dx

= 2x^(1/2)e^√x - ∫ e^√x/x^(1/2) dx

= 2x^(1/2)e^√x – 2e^√x

2 ∫ x^(1/2)e^√x dx

= 2 ( 2xe^√x – 2 ∫ 2x^(1/2)e^√x/(2x^(1/2)) dx )

= 2 [ 2xe^√x – 2 ( 2x^(1/2)e^√x – 2e^√x ) ] + C2

= 4 e^√x ( x – 2x^(1/2)+ 2 ) + C2

∫ e^(2√x) dx

令 u = x^(1/2), dv = e^(2√x)/x^(1/2) dx

則 du = 1/(2x^(1/2)) dx, v = e^(2√x)

∫ e^(2√x) dx

= x^(1/2)e^2√x - ∫ 1/2 * e^(2√x)/x^(1/2) dx

= x^(1/2)e^2√x – 1/2 * e^(2√x) + C3

= 1/2 * e^2√x ( 2x^(1/2) – 1 ) + C3

三式合併

∫ ( x^(1/2) + e^√x )^2 dx

= ∫ x dx + 2 ∫ x^(1/2)e^√x dx + ∫ e^(2√x) dx

= x^2/2 + 4 e^√x ( x – 2x1/2 + 2 ) + 1/2 * e^2√x ( 2x^(1/2) – 1 ) + C

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∫ x^3 + 1/x dx

= x^4/4 + ㏑|x| + C

這樣打, 可能會有點搞糊塗, 建議到下列網址看答案

http://tw.myblog.yahoo.com/jw!JCVbjQyaBRbXTWOakincl1.Wpxbobg--/article?mid=1351&prev=582&next=1350

恩恩沒錯沒錯
也謝謝你的作答喔

第2題是要積 x/(x^3+1)?

做部分分式分解, 得
　 x/(x^3+1) = -(1/3)/(x+1) + (1/3)(x+1)/(x^2-x+1)
故
　 ∫x/(x^3+1) dx = -(1/3)∫1/(x+1) dx + (1/3)∫(x+1)/(x^2-x+1) dx
　　　　　　　= (-1/3)ln|x+1| + (1/3)∫(x-1/2+3/2)/(x^2-x+1) dx
　　　　　　　 = (-1/3)ln|x+1| +(1/6)ln|x^2-x+1|+(1/2)∫1/[(x-1/2)^2+3/4] dx