請問這幾題不定積分怎麼解?

2016-04-09 10:15 am
1.(t^3)(1+t)^101
2.[(t-1)^5](3-t)^3
3.(x^2+3x+4)/(x^2-5x+6)

回答 (2)

2016-04-09 11:55 am
✔ 最佳答案
請看:
http://s18.postimg.org/7341u73xz/image.png
2016-04-19 4:17 am
1.
I=積分(t^3)(1+t)^101 dt
If u=1+t
t=u-1
du=dt
I=積分(u-1)^3 t^101 du
I=積分(u^3 -3u^2+3u+1)(u^101) du
I=積分(u^104 -3u^103+3u^102 +u^101) du
I=(u^105)/105-(u^104)/104+(3u^103)/103+(u^102)/102 +C
I=(1/105)(1+t)^105-(1/104)(1+t)^104+(1/103)(1+t)^103+(1/102)(1+t)^102+C

2.
積分[(t-1)^5](3-t)^3 dt
=積分(t^5 -t^4 +10t^3 -10t^2 +5t -1)(27-9t+3t-t^3)dt
=積分(27t^5 -9t^6+3t^7 -t^8 -135t^4+45t^5-15t^6+5t^7+270t^3-90t^4+30t^5-10t^6+135t-45t^2+15t^3-5t^4-27+9t-3t^2+t^3) dt
=積分(-t^8+8t^7 -34t^6 +112t^5 -260t^4+376t^3 -318t^2+144t-27)dt
=-(t^9)/9+t^8 -(34/7)t^7+(56/3)(t^6)-52t^5 +94t^4 -106t^3+72t^2-27t+C

3.
積分 (x^2+3x+4)/(x^2-5x+6) dx
=-14log(x-2)+22log(x-3)+x+C


收錄日期: 2021-04-18 14:42:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160409021535AAEs2kA

檢視 Wayback Machine 備份