1.int (e^x/((e^x-2)(e^x+6))dx
2.int (3x^2-4x-1)/(x^3-2x^2-x+2)dx hint: (3x^2-4x-1)/(x^3-2x^2-x+2 = 1/(x+1)+1/(x-1)+1/(x-2).
I have answered Q1= 1/8(ln(2-e^x)-ln(e^x+6))+C and Q2=ln(x+1)+ln(x-1)+ln(x-2)+C,
but webwork responsed that " there is always more than one possibility"
Plz help!
two integration problems
2014-03-24 7:23 pm
回答 (2)
2014-03-24 7:58 pm
✔ 最佳答案
Please read:圖片參考：https://s.yimg.com/rk/HA00430218/o/104416062.png
2014-03-24 15:37:37 補充：
I agree.
^____^
2014-03-24 11:33 pm
For Q2, should we need to use the hints? (Otherwise, no marks will be given?)
For me, I will do in this way,
As (3x^2 - 4x - 1)dx = d(x^3 - 2x^2 - x + 2), so,
∫ [(3x^2 - 4x - 1)/(x^3 - 2x^2 - x + 2)] dx
= ∫ [1/(x^3 - 2x^2 - x + 2)] d(x^3 - 2x^2 - x + 2)
= ln|x^3 - 2x^2 - x + 2| + C
For me, I will do in this way,
As (3x^2 - 4x - 1)dx = d(x^3 - 2x^2 - x + 2), so,
∫ [(3x^2 - 4x - 1)/(x^3 - 2x^2 - x + 2)] dx
= ∫ [1/(x^3 - 2x^2 - x + 2)] d(x^3 - 2x^2 - x + 2)
= ln|x^3 - 2x^2 - x + 2| + C
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