Re: [積分] 我想問兩題INTEGRATION BY PARTS
※ 引述《tomho1202 (never regreat)》之銘言:
: x(e)^2x 1 e^2x
: ∫————————dx = --- ------- +c
: (2x+1)^2 4 (2x+1)
2x 2x
d e 4xe
--- ---------- = ---------
dx 2x+1 (2x+1)^2
: ∫ x(x-1)^1/2 dx
2
=∫ (u +1 ) u *2udu
4 2
=∫2u + 2u du = (2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + c
let u= (x-1)^(1/2)
2udu =dx
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