1樓：滾雪球的祕密
解答：x^2+2x+3 = (x+1)^2 +2
letx+1 =√2tanu
dx = √2(secu)^2 .du
∫ (x+3)/(x^2+2x+3)^2 dx
=(1/2) ∫ (2x+2)/(x^2+2x+3)^2 dx + 2∫ dx/(x^2+2x+3)^2
=-(1/2) [1/(x^2+2x+3)] +2∫ dx/(x^2+2x+3)^2
=-(1/2) [1/(x^2+2x+3)] +2∫ √2(secu)^2 .du/[ 4(secu)^4]
=-(1/2) [1/(x^2+2x+3)] +(√2/2)∫ (cosu)^2 du
=-(1/2) [1/(x^2+2x+3)] +(√2/4)∫ (1+cos2u) du
=-(1/2) [1/(x^2+2x+3)] +(√2/4)[ u +(1/2)sin2u] +c
=-(1/2) [1/(x^2+2x+3)] +(√2/4) +c
x+1 =√2tanu
sinu = (x+1)/√(x^2+2x+3)
cosu =√2/√(x^2+2x+3)
2樓：道清逸森君
∫(x+3)/(x²+2x+10)
dx=∫
(x+1+2)/(x²+2x+10)
dx=(1/2)∫
(2x+2)/(x²+2x+10)dx+
2∫1/(x²+2x+10)
dx=(1/2)∫
1/(x²+2x+10)
d(x²+2x)+2∫
1/[(x+1)²+9]
dx=(1/2)ln(x²+2x+10)
+(2/3)arctan[(x+1)/3]+c【數學之美】團隊為您解答，若有不懂請追問，如果解決問題請點下面的“選為滿意答案”。
3樓：
這道題，，，有點複雜
求（x－1）／（x^2+2x+3）的不定積分
4樓：不是苦瓜是什麼
|^∫(x-1)/(x²+2x+3)dx
=½∫(2x-2)/(x²+2x+3)dx
=½∫(2x+2-4)/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx
=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx
=½ln|x²+2x+3| - ∫1/dx + c
=½ln|x²+2x+3| - (√2)∫1/d[(x+1)/√2] + c
=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + c
不定積分的公式
1、∫ a dx = ax + c，a和c都是常數
2、∫ x^a dx = [x^(a + 1)]/(a + 1) + c，其中a為常數且 a ≠ -1
3、∫ 1/x dx = ln|x| + c
4、∫ a^x dx = (1/lna)a^x + c，其中a > 0 且 a ≠ 1
5、∫ e^x dx = e^x + c
6、∫ cosx dx = sinx + c
7、∫ sinx dx = - cosx + c
8、∫ cotx dx = ln|sinx| + c = - ln|cscx| + c
9、∫ tanx dx = - ln|cosx| + c = ln|secx| + c
5樓：基拉的禱告
詳細過程如圖所示，令x+1=t換元做，希望對你有所幫助，望採納哦
6樓：體育wo最愛
||令x=t²，dx=2tdt
原式=∫[2t/(1+t³)]dt=2∫[t/(1+t)(1-t+t²)]dt
=(2/3)∫[(1+t)/(1-t+t²)-1/(1+t)]dt
=(-2/3)ln|1+t|+(1/3)∫[(2t+2)/(t²-t+1)]dt
=(-2/3)ln|1+t|+(1/3)∫[(2t-1)+3]/(t²-t+1)dt
=(-2/3)ln|t+1|+(1/3)∫[(2t-1)/(t²-t+1)]+∫[1/(t²-t+1)]dt
=(-2/3)ln|t+1|+(1/3)∫[1/(t²-t+1)]d(t²-t+1)+∫[1/(t-1/2)²+(√
3/2)²]dt
=(-2/3)ln|t+1|+(1/3)ln(t²-t+1)+(2/√3)arctan[(2t-1)/√3]+c
將t=√x代入上式即得
7樓：匿名使用者
^令w=x^1/6
則x=w^6,dx=6w^5dw
則原式=6∫w^3/(w+1)dw=6∫(w^3+1-1)/(w+1)dw
=6∫[(w^2-w+1)-1/(w+1)]dw=2w^3-3w^2+6w-ln(w+1)+c
帶入w=x^1/6
得原式=2x^1/2-3x^1/3+6x^1/6-ln(1+x^1/6)+c
樓上的代換形式也是正確的，但在中間計算過程中可能有錯誤。
8樓：匿名使用者
|∫[1/(x²-2x-3)]dx
=∫[1/(x+1)(x-3)]dx
=¼∫[(x+1)-(x-3)]/[(x+1)(x-3)] dx=¼∫[1/(x-3) -1/(x+1)]dx=¼∫[1/(x-3)]d(x-3) -¼∫[1/(x+1)]d(x+1)
=¼ln|x-3|-¼|ln(x+1)|+c=¼ln|(x-3)/(x+1)| +c
9樓：匿名使用者
1/(x^2-2x-3) = (1/4)[1/(x-3) -1/(x+1)]
∫dx/(x^2-2x-3)
=(1/4)∫[1/(x-3) -1/(x+1)] dx=(1/4) ln|(x-3)/(x+1)| + c
10樓：別問
^換元法，令w=1+x^1/6
得到化簡後
原式積分=\int 6w-12+6/w dw=3w^2 -12w + 6 log(w) + c代換回來即得到
積分=x^1/3 - 6x^1/6 + 6log(1+x^1/6) + c
11樓：匿名使用者
^原式=∫dx/((x+1)^2+2)^2x+1=√2tanu sin2u=2√2(x+1)/(x^2+2x+3)
=∫√2(secu)^2du/[4(secu)^4]=(√2/8)∫(1+cos2u)du
=√2u/8+√2sin2u/16
=(√2/8)arctan[(x+1)/√2]+(x+1)/[4(x^2+2x+3)]+c
12樓：綠意如煙
∫(x-1)/(x²+2x+3)dx =½∫(2x-2)/(x²+2x+3)dx =½∫(2x+2-4)/(x²+2x+3)dx =½∫...
13樓：懶懶的小杜啦
|∫x3/(x2+2x-3)dx=∫(x3+2x-3x-2x+3)/(x2+2x-3)dx =∫x+3/(x2+2x-3)dx =∫xdx+3∫1/(x2+2x-3)dx =x2/2+3∫1/[(x-1)(x+1)]dx =x2/2+3/4∫1/(x-1)-1/(x+3)]dx = x2/2+3/4ln|x-1|-3/4ln|x+3|+c
14樓：匿名使用者
我想問一下第三步的後面一部分怎麼解的
15樓：孤狼嘯月
原式＝∫
（x+1-2）/（x²+2x+3）dx
＝∫（x²/2+x）/（x²+2x+3）dx-∫2/[2+（1+x）²]dx
＝1/2*ln（x²+2x+3）-∫1/[1+（1/✓2 +x/✓2）²]dx
＝1/2*ln（x²+2x+3）-✓2*arctan（x/✓2+1/✓2）+c
∫（x-1）/（x^2+2x+3）dx的不定積分怎麼求
16樓：安克魯
|∫(x-1)/(x²+2x+3)dx
=½∫(2x-2)/(x²+2x+3)dx
=½∫(2x+2-4)/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx
=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx
=½ln|x²+2x+3| - ∫1/dx + c
=½ln|x²+2x+3| - (√2)∫1/d[(x+1)/√2] + c
=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + c
17樓：無敵粥
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