## elica85 2 years ago integrate...

1. elica85

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

2. elica85

take V,m, and M as constants..

3. zepdrix

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? :)

4. elica85

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

5. elica85

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

6. elica85

but i don't

7. zepdrix

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

8. elica85

sorry i will redo with different letters...

9. elica85

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

10. elica85

d=c-bt c=e+f

11. zepdrix

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

12. elica85

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

13. zepdrix

Yah looks good so far c:

14. elica85

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

15. zepdrix

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

16. zepdrix

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

17. elica85

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

18. elica85

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

19. zepdrix

yay c: