indefinite integration urgent!

Katherine

2011-09-01 6:41 am

http://imageshack.us/photo/my-images/854/39456138.png/
請唔好跳步驟因為我係蠢人所以唔該各位高手有咁齊寫咁齊
間中有d解釋仲好(e.g 用左邊條rule)
thx!!

回答 (1)

CRebecca

2011-09-01 8:42 am

✔ 最佳答案

1. Let u= 2x-3, x=(u+3)/2, dx= (1/2) du
∫ x/(3-2x)² dx
= (1/4)∫ (u+3)/u² du
=(1/4) [ ln|u|- 3/u ]+ c
=(1/4)[ ln| 2x-3 | - 3/(2x-3) ] +c

2. Let u=x^4, du= 4x^3 dx, x^11 dx= (1/4) u² du
∫ x^11 /(x^8+ 3x^4+2) dx
=(1/4) ∫ u²/(u²+3u+2) du
=(1/4) ∫ [ 1 + 1/(u+1) - 4/(u+2) ] du
=(1/4) [ u + ln(u+1)- 4ln(u+2) ] +C
=(1/4) [ x^4+ ln(x^4 +1) - 4 ln(x^4+ 2) }+c

3.∫ dx/[x(1+x^5)]
=∫ dx/[ x^6 ( 1+ 1/x^5 ) ] (let u=1+ 1/x^5= x^(-5), du= -5/x^6 dx )
=(-1/5) ∫ du/ u
=(-1/5) ln|u| +c
=(-1/5) ln| 1+ 1/x^5 | +c
or ln|x| - (1/5) ln| x^5 + 1 | +c

2011-09-10 02:41:50 補充：
Q3: [let u=1+ 1/x^5=1+ x^(-5), du= -5/x^6 dx ]

👍 0 👎 0