討論串[積分]
共 259 篇文章
內容預覽:
= sin^-1 x* (1/2)x^2 -∫(1/2)x^2 * 1/(1-x^2)^(1/2)dx. set (1-x^2)^(1/2)=u 感謝糾正. 1-u^2=x^2. 2xdx = -2udu. = sin^-1 x* (1/2)x^2 -(1/2)∫(1-u^2)^(-1/2)du.
(還有321個字)
內容預覽:
∫ln [(x+1)^2 + 1] dx. set x+1 =tanθ. dx =(secθ)^2 dθ. ∫2 ln secθ*(secθ)^2 dθ. 分部積分. =2[ln secθ*tanθ -∫(tanθ)^2 dθ] +c. =2[ln secθ*tanθ - tanθ+ θ] +c.
(還有28個字)