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!

Anonymous2012-06-08T14:48:47Z

Favorite 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

MechEng20302012-06-08T21:34:53Z

∫v/(9.8 - v) dv => -∫v/(v - 9.8)) dv => -∫(v - 9.8 + 9.8)/(v - 9.8) dv => -∫(1 + 9.8/(v - 9.8)) dv

= -v - 9.8*ln|v - 9.8| + C

Iggy Rocko2012-06-08T21:28:44Z

..................-1
............---------
-v + 9.8 | v + 0
.............v - 9.8
.............---------
................9.8

v/(9.8 - v) = -1 + 9.8/(9.8 - v)
∫ v/(9.8 - v) dv =
∫ -1 + 9.8/(9.8 - v) dv =
-v - 9.8ln(9.8 - v) + c