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how to integrate this improper fraction?
integrate v / (9.8-v) dv
I thought the answer was -v ln(9.8-v) but that doesn't seem right.. Please help and explain!
Cheers!
3 Answers
- Anonymous9 years agoFavorite Answer
I would change my variable:
A=9,8-v
dA=-dv
So i'd have:
"integrate [(-A+9,8)/A](-dA)"=
="integrate [(A-9,8)/A]dA"="integrate[(A/A)-(9,8/A)]dA"=
="integrate [1-(9,8/A)]dA"=
="integrate [1]dA"-"integrate (9,8/A)dA"=
=A-9,8*ln(A)+C
Change back:
=9,8-v-9,8*ln(9,8-v)+C
Put the first term into C:
=-v-9,8*ln(9,8-v)+K
- MechEng2030Lv 79 years ago
â«v/(9.8 - v) dv => -â«v/(v - 9.8)) dv => -â«(v - 9.8 + 9.8)/(v - 9.8) dv => -â«(1 + 9.8/(v - 9.8)) dv
= -v - 9.8*ln|v - 9.8| + C
- Iggy RockoLv 79 years ago
..................-1
............---------
-v + 9.8 | v + 0
.............v - 9.8
.............---------
................9.8
v/(9.8 - v) = -1 + 9.8/(9.8 - v)
â« v/(9.8 - v) dv =
â« -1 + 9.8/(9.8 - v) dv =
-v - 9.8ln(9.8 - v) + c