x1x22x3dx的不定積分怎麼求

2021-03-04 09:01:18 字數 3841 閱讀 5196

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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