Re: [積分] 多項式分式
※ 引述《mmmbop (勇敢)》之銘言:
: 2
: x - 3x + 1
: ∫ _________________ dx
: 2
: (x+4)(x - 6x + 13 )
: 想很久了算不出來 Orz 拜託大俠了
A B x + C
原式 = S _____ + _____________ dx
x + 4 2
( x -6x +13)
2 2
A( x -6 x +13)+(Bx+C)(x + 4) = x - 3x +1
展開先比較系數,得 A+B= 1 -6A+4B+C = -3 13A+4C = 1====>可解3個未知數
得到 A =29/53 B =24/53 C = -81/53 ==>建議用克拉馬去解,或用代入消去,要有耐心
才解得出來
29/53 24/53 x -81/53 29/53 (24/53 x-72/53)- 9/53
代入,得 S _____ + _______________ dx = S ______ + _____________________ dx
x + 4 2 x + 4 2 2
( x -6 x +13 ) ( x -3 ) + 2
1 ( x - 3 ) 1
= 29/53 S ______ dx + 24/53 S ______________ dx -9/53 S __________________ dx
x + 4 2 2 2 2
( x - 3) + 2 ( x -3 ) + 2
2 -1 x - 3
=29/53 ln (x + 4) + 12/53 ln ( x - 3 ) - 9/53 { 1/2 tan ( ________ )} + C
2
-1 x - 3
=29/53 ln (x + 4 ) + 24/53 ln ( x - 3 ) - 9/106 tan ( _________ ) + C ##
2
算是還好,但是要po很久...><",這題需要注意我上面所用到的小技巧,希望對你有些幫助
,有問題盡量po版,加油吧.............^^.....
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.114.212.82
→
140.114.212.82 01/05, , 1F
140.114.212.82 01/05, 1F
推
218.34.226.161 01/05, , 2F
218.34.226.161 01/05, 2F
推
218.34.226.161 01/05, , 3F
218.34.226.161 01/05, 3F
→
218.34.226.161 01/05, , 4F
218.34.226.161 01/05, 4F
推
140.114.212.82 01/05, , 5F
140.114.212.82 01/05, 5F
推
140.114.212.82 01/05, , 6F
140.114.212.82 01/05, 6F
推
218.34.226.161 01/05, , 7F
218.34.226.161 01/05, 7F
推
140.114.212.82 01/05, , 8F
140.114.212.82 01/05, 8F
→
140.114.212.82 01/05, , 9F
140.114.212.82 01/05, 9F
→
140.114.212.82 01/05, , 10F
140.114.212.82 01/05, 10F
推
218.34.226.161 01/05, , 11F
218.34.226.161 01/05, 11F
→
218.34.226.161 01/06, , 12F
218.34.226.161 01/06, 12F
推
140.114.212.82 01/06, , 13F
140.114.212.82 01/06, 13F
討論串 (同標題文章)