Question

integrate...

Answer 1

\[\int\limits_{0}^{t} \frac{ Vm }{ M-mt } dt\]

Answer 2

take V,m, and M as constants..

Answer 3

It might be a little easier to read if pull the constants (in the numerator) outside of the integral, and then apply a `U substitution`.\[\large \int\limits\limits_{0}^{t} \frac{Vm}{M-mt} dt \qquad \rightarrow \qquad Vm\int\limits\limits_{0}^{t} \frac{1}{M-mt}dt\]

Are you familiar with U substitution? :)

Answer 4

thx, yes, what i get is -V{ln[(M-mt)/M]}...

Answer 5

now if M1=M-mt,
M=Mb+Mf...and you substitue and simplify, i'm suppose to get Vln[1+(Mf/Mb)]

Answer 6

but i don't

Answer 7

I'm so confused, there are too many M's floating around :( lol

Answer 8

sorry i will redo with different letters...

Answer 9

\[\int\limits_{0}^{t} \frac{ ab }{ c-bt } dt\]
all constants except t

Answer 10

d=c-bt
c=e+f

Answer 11

So integrating, before doing any fancy simplification should give you,\[\large ab\left(-\frac{1}{b}\right)\ln|c-bt|_0^t\]

Moar letters? :U oh boy lol

Answer 12

ok so integrate over limits and it should be -a[ln(c-bt)-ln(c)]
correct so far?

Answer 13

Yah looks good so far c:

Answer 14

which is -a{ln[(c-bt)/c]}

Answer 15

Yes, you can bring the negative into the log as an exponent. Which will give us this,\[\large a \ln\left[\left(\frac{c-bt}{c}\right)^{-1}\right] \qquad = \qquad a \ln\left[\left(\frac{c}{c-bt}\right)\right]\]

Whichever way makes sense to you though :3

Answer 16

Now what's going on with this d and e and f nonsense? XD lol

Answer 17

ok, so f has to be in there cuz that's what i'm trying to solve. so plugging in the other letters and simplifying, according to the solution, it's suppose to be
a[ln(1+(f/e))] but i can't get it. however, i haven't tried with your second expression up there so i will try it now

Answer 18

ok, i think one of the variables is messed up on the solution. i got it, thx for your help

Answer 19

yay c: