Question

Integral of 1/t^2(t-1) dt

Answer 1

Is the answer \[\left(\begin{matrix}1 \\ t\end{matrix}\right) - \ln \left(\begin{matrix}t \\ t-1\end{matrix}\right) + c\] the same as
\[\frac{ -2 }{ t^3(t-1) } + \frac{ \ln(t-1) }{ t^2 } + c\]

Answer 2

I used t^2 = u and dv = 1/(t-1)

Answer 3

What is the integral? \(\int\frac{1}{t^2(t-1)}dt\)?
What is \(\binom{1}{t}\) and \(\ln\binom{t}{t-1}\)? Binomial coefficients?

Answer 4

the correct answer is \[\frac{ 1 }{ t } - \ln \frac{ t }{ t-1 }\]
and what I have written as the last equation is my solution

Answer 5

and yes thats teh integral

Answer 6

you did integration by parts ?

Answer 7

when both the functions of of same type, here algebraic,
integration by parts is not recommended.

for this problem, partial fractions is the easiest way.
know what partial fractions is ??

Answer 8

yes I made u = t^-2 and dv = (t-1)^.-1= 1(/(t-1)

Answer 9

I have no idea how to tackle this integral. \(t=\sec^2(x)\)?

Answer 10

I don't by terms probably, but i've probably done it, could u explain?

Answer 11

sure
\(\huge \dfrac{1}{t^2(t-1)}=\dfrac{A}{t}+\dfrac{B}{t^2}+\dfrac{C}{t-1}\)

you just need to find A,B,C

Answer 12

Oh yes!, I have done this, okey let me try to see if I can solve this.

Answer 13

take your time :)

Answer 14

okey so A = 2 B and C = -1?

Answer 15

doesn't seem to be correct, how did u get them ?

Answer 16

1 = At (t-1) +B (t-1) +C (t^2)

plug in t=0, you will get B
plug in t= 1, you will get C

Answer 17

-b = 1 that gives b = -1.
Sorry first of the equation 1 = A(t()t-1) + b(t-1) + c(t^2)
and now after getting b, I have 1 = -At -Bt and if t = 1 u have 1 = -A-B so -A = 1+B hat gives A = -1-B= -1-(-1) = 0 so I was wrong here okey, and C, 1 = At^2+ Ct^2 and if t=1 then 1 = A+C which gives C equals to 1, I was wrong here again.
Conclusion, A = 0, B = -1, C = 1

Answer 18

how can u say -b=1 ?

Answer 19

B has a term that doesnt include t B(t-1) which is Bt and -B, but you way of doing this was easier.

Answer 20

oh yes!, B=-1 is correct :)
find A and C now

Answer 21

C= 1 is also corect

Answer 22

but A is Not 0

Answer 23

A =-1

Answer 24

when t = 2

Answer 25

But the answer in my book is \[\frac{ 1 }{ t } - \ln \frac{ t }{ t-1 } + c\]

Answer 26

you will get that :)
plug in A,B,C values!

Answer 27

\[\frac{ -1 }{ t }- \frac{ 1 }{ t^2 }\ + \frac{ 1 }{ t-1 }\]

Answer 28

and now?

Answer 29

yes
now integrate!

Answer 30

-1/t^2 is -t^-2
its integral will be ?

Answer 31

oh okey, 1 sec

Answer 32

-t^-1 = -1/(t)

Answer 33

thats integral of just t^(-2)

Answer 34

okey hold on

Answer 35

t^-1 = 1/(t)

Answer 36

yes, correct
and what about other 2 terms, you can integrate those ?

Answer 37

yes, give me a sec:)

Answer 38

okey so for -t^-1 the answer is -ln t
for 1/(t-1) its ln (t-1)?

Answer 39

correct

Answer 40

factor out the negative

-ln t + ln (t-1) = - [ ln t - ln (t-1)]

Answer 41

= - ln [t/(t-1)]
:)

Answer 42

God... I was thinking how to use trig substitution for \(\int\frac{1}{t^2}dt\)...

Answer 43

oh right thats how you do it, I gotta tell you ever since I have progressed in the math courses I've started to felt more stupid, is that normal?

Answer 44

feeö*

Answer 45

feel

Answer 46

ofcourse...you're learning, silly mistakes are bound to happen, and thats a good thing! :)

Answer 47

Well, huge thanks to both of you and for taking ur time ;)

Answer 48

welcome ^_^
always happy to help :)

Answer 49

hehe, well you will probably see more of me later or sooner, so I'll go ahead and close this thread now:)