1樓:安克魯
|∫(x-1)/(x2+2x+3)dx
=1⁄2∫(2x-2)/(x2+2x+3)dx
=1⁄2∫(2x+2-4)/(x2+2x+3)dx
=1⁄2∫(2x+2)/(x2+2x+3)dx - 1⁄2∫4/(x2+2x+3)dx
=1⁄2∫(2x+2)/(x2+2x+3)dx - 2∫1/(x2+2x+3)dx
=1⁄2∫d(x2+2x+3)/(x2+2x+3) - 2∫1/[(x+1)2+2]dx
=1⁄2ln|x2+2x+3| - ∫1/dx + c
=1⁄2ln|x2+2x+3| - (√2)∫1/d[(x+1)/√2] + c
=1⁄2ln|x2+2x+3| - (√2)arctan[(x+1)/√2] + c
2樓:無敵粥
分母下變成(x-1)(x+3). ∫(x-1)/(x^2+2x+3)dx=∫1/(x+3)dx =ln(x+3)
求(x-1)/(x^2+2x+3)的不定積分
3樓:不是苦瓜是什麼
|^∫(x-1)/(x2+2x+3)dx
=1⁄2∫(2x-2)/(x2+2x+3)dx
=1⁄2∫(2x+2-4)/(x2+2x+3)dx
=1⁄2∫(2x+2)/(x2+2x+3)dx - 1⁄2∫4/(x2+2x+3)dx
=1⁄2∫(2x+2)/(x2+2x+3)dx - 2∫1/(x2+2x+3)dx
=1⁄2∫d(x2+2x+3)/(x2+2x+3) - 2∫1/[(x+1)2+2]dx
=1⁄2ln|x2+2x+3| - ∫1/dx + c
=1⁄2ln|x2+2x+3| - (√2)∫1/d[(x+1)/√2] + c
=1⁄2ln|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
4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
5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
6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。
7樓:匿名使用者
|∫[1/(x2-2x-3)]dx
=∫[1/(x+1)(x-3)]dx
=1⁄4∫[(x+1)-(x-3)]/[(x+1)(x-3)] dx=1⁄4∫[1/(x-3) -1/(x+1)]dx=1⁄4∫[1/(x-3)]d(x-3) -1⁄4∫[1/(x+1)]d(x+1)
=1⁄4ln|x-3|-1⁄4|ln(x+1)|+c=1⁄4ln|(x-3)/(x+1)| +c
8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
9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
10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
11樓:綠意如煙
∫(x-1)/(x2+2x+3)dx =1⁄2∫(2x-2)/(x2+2x+3)dx =1⁄2∫(2x+2-4)/(x2+2x+3)dx =1⁄2∫...
12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
13樓:匿名使用者
我想問一下第三步的後面一部分怎麼解的
14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^2+2x+1)dx/(x^3+x^2+x+1)求這個不定積分的方法步驟過程,謝謝啦 20
15樓:匿名使用者
|分母因式分解為:(x+3)(x-1)
令:(2x+1)/[(x+3)(x-1)]=a/(x+3)+b/(x-1)
右邊通分合併,與左邊比較係數後得:a=5/4,b=3/4則:∫ (2x+1)/(x2+2x-3) dx=(5/4)∫ 1/(x+3) dx + (3/4)∫ 1/(x-1) dx
=(5/4)ln|x+3| + (3/4)ln|x-1| + c
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