不定積分 使用代換法

1. u=tan(x/2), x= 2arctan(u), dx= 2du/(1+u^2), sinx= 2u/(1+u^2)
∫dx/(1+sinx)=∫(1+u^2)/(1+u)^2 * 2du/(1+u^2)
=∫2du/(1+u)^2 = -2/(1+u) + C
= -2/[1+tan(x/2)]+ C

2.
設 f(x)=∫(x^2+1)/(x^4+1) dx
g(x)=∫(x^2-1)/(x^4+1) dx
(1) f(x)=∫(1+ 1/x^2)/[(x-1/x)^2 + 2] dx , 令u=x-1/x
=∫ du/(u^2+2) = 1/√2 *arctan(u/√2) + C
= 2/√8 * arctan[(x^2-1)/(x√2)]+ C
(2) g(x)=∫(1- 1/x^2)/[(x+1/x)^2 - 2] dx , 令 u= x+ 1/x
=∫ du/(u^2 - 2) = 1/(2√2) * ln| (u-√2)/(u+√2) | + C
= 1/√8 ln| (x^2-x√2+1)/(x^2+x√2 +1) | + C
原題=∫dx/(x^4+1)= [f(x)-g(x)]/2
={ 2arctan[(x^2-1)/(x√2)] - ln|(x^2-x√2+1)/(x^2+x√2+1)| }/(4√2) + C
∫∫∫∫∫∫

謝謝doraemonpaul 讓我看到詳細過程

2.∫dx/(x^4+1)
http://hk.knowledge.yahoo.com/question/question?qid=7008072801998

第二題:
留意x^4+1=(x^2+1)^2-(√2*x)^2=(x^2+1-√2*x)(x^2+1+√2*x)
再用partial fraction,先後利用公式∫dx/(x^2+a^2)=(1/a)(arctan(x/a)+c

據教據教育部98新課程高三下學期不定積分

題目：∫dx/(1+sinx)
提示：let u = tan(x/2)
【作法】
由u = tan(x/2)
得sin(x/2) = u/[√(1+u²)] , cos(x/2) = 1/[√(1+u²)]
故sin(x) = 2sin(x/2)cos(x/2) = 2u/(1+u²)
又由連鎖律(Chain-Rule)、變數變換和三角函數公式得
du = sec²(x/2)乘(½)dx = [1+tan²(x/2)]乘(½)dx = (1+u²)乘(½)dx
故原式∫dx/(1+sinx) = ∫[(du/(½)(1+u²)]/{1+[2u/(1+u²)]} = ∫2du/(u²+2u+1) = 2∫du/(1+u)² = -2(1+u)^(-1) +C
故所求即為-2[1+tan(x/2)]^(-1)+C
【另類想法】
∫dx/(1+sinx) = ∫(1-sinx)dx/(1+sinx)(1-sinx) = ∫(1-sinx)dx/(1-sin²x) = ∫(1-sinx)dx/cos²x = (∫dx/cos²x) – (∫sinxdx/cos²x )= ∫sec²xdx - ∫tanxsecxdx = tanx-secx+C

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∫dx/(x^4+1)
= [(√2)/8]ln|[x²+(√2)x+1]/[x²-(√2)x+1]|+ (√2)/4{tan^(-1)[(2x+√2)/ (√2)]+tan^(-1) [(2x-√2)/ (√2)]}+C

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雖盡力答覆恐尚有疏漏，期望各界不吝指正與提供更不一樣的想法、不以抄襲，謝謝您的合作