推 noyl91 :感謝你~ 05/27 20:45
※ 引述《noyl91 (azure)》之銘言:
: 1 6x^2 - 3x + 1
: ∫ ---------------- dx
: 0 (4x+1)(x^2+1)
: 試過[lnf(x)]'跟變數代換(令x=tan日)
: 但還是解不出來...
: 還請各位幫忙給個方向,
: 謝謝!
: Ans: 1/2[ln10 - pi/2 ]
6x^2 - 3x + 1 A Bx + C A(x^2+1) + (Bx+C)(4x+1)
------------- = ------ + ------- = -----------------------
(4x+1)(x^2+1) 4x + 1 x^2 + 1 (4x+1)(x^2+1)
A + 4B = 6 ---> B + 7 + 16B = 24 = 17B + 7
B + 4C = -3 ---> B + 7 = 4A
A + C = 1 ---> C = 1 - A
A = 2, B = 1, C = -1
剩下的自己解
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