有哪位微积分好的大哥帮忙解一下这个式子吗？越快越好，求求

Question

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Answer 1

题目要求你解析解了？感觉解不出来啊

Answer 2

计算过程很繁。主要步骤是 ：被积函数分子分母同除以 √(2+x).
令 √[(4-x)/(2+x)] = t, 得 x = (4-2t^2)/(1+t^2) = 6/(1+t^2) - 2
dx = -12tdt/(1+t^2)^2.
I = ∫<0, 2>dx/{√[(4-x)/(2+x)]+1} = ∫<1/√2, √2>12tdt/[(1+t)(1+t^2)^2]
化部分分式
I = 3∫<1/√2, √2>[ (2t+2)/(t^2+1)^2 + (t-1)/(t^2+1) - 1/(t+1)]dt
前项令 t = tanu
I = 3∫<arctan(1/√2), arctan√2>(sin2u+cos2u+1)du
+3 [(1/2)ln(t^2+1) - arctant - ln(t+1)]<1/√2, √2>
= 3[u+(1/2)sin2u-(1/2)cos2u]<arctan(1/√2), arctan√2>
+ 3[arctan(1/√2) - arctan√2]
= 3[u+sinucosu-(cosu)^2+1/2)]<arctan(1/√2), arctan√2>
+ 3[arctan(1/√2) - arctan√2]
= 3[arctan√2 - arctan(1/√2) +1/3]+ 3[arctan(1/√2) - arctan√2] = 1本回答被提问者采纳