Solve Integral of t/1+t^2 - 1/2 ln (1+t^2) + c

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Homework Help Overview

The discussion revolves around the integral of the function t/(1+t^2) with respect to t. Participants are comparing their answers and exploring the correct method for integration, particularly focusing on substitution techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their attempts at integration, with one expressing confusion over their answer compared to a referenced solution. Others inquire about the methods used, particularly substitution, and question the steps involved in the integration process.

Discussion Status

Some participants have provided partial calculations and corrections, while others are seeking clarification on specific steps. There is an ongoing exploration of different interpretations of the integral and the substitution method.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There is uncertainty regarding the correctness of various steps in the integration process, and assumptions about the variables involved are being questioned.

rojak44
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hello there,im having trouble with this question..
find int t/1+t^2 dt

my answer is 1/2 ln t (1+t^2)+c
but the answer that i have copy is 1/2 ln (1+t^2)+c
maybe i have copy it wrongly,can someone tell me which one is right?
thx.. :smile:
 
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How did you get your answer?
 
im using the integration using substitution..
im confuse right now.. can anyone give me the full calculation or what method i should use.
 
\frac{t}{1+t^2} dt

Let u=1+t^2 what is \frac{du}{dt}? then what is dt in terms of du?
 
here it is,pls correct me if I am wrong.
int t/1+t^2 dt
u = 1+t^2
du/dt = 2t
dt = 1/2t du
= int (t/u)(1/2t du)
= int t/2tu du
= 1/2 int t/tu du
= 1/2 int t(tu)^-1 du
= 1/2 int t^0 u^-1 du
= 1/2 [t^1 ln u]+c
= 1/2 ln t (1+t^2)+c (answer)
 
Last edited:
could anyone help me here? pls..
 
rojak44 said:
here it is,pls correct me if I am wrong.
int t/1+t^2 dt
u = 1+t^2
du/dt = 2t
dt = 1/2t du
= int (t/u)(1/2t du)
= int t/2tu du
= 1/2 int t/tu du
= 1/2 int t(tu)^-1 du
= 1/2 int t^0 u^-1 du
= 1/2 [t^1 ln u]+c
= 1/2 ln t (1+t^2)+c (answer)
Cancel out t in this step.
 
u mean remove the t? so it will become 1/2u?
here,
= int 1/2u du
= int u^-1/2 du
= 1/2 int u^-1 du
=1/2 [ln u]+c
= 1/2 ln (1+t^2)+c (answer)
am i correct? btw if we remove t,it will become 1/2u right since it was 1t/2tu?
 
Yes, that is correct.
 
  • #10
thank you! :)
i got one more question,
differentiate y= 2m e^mt/cos 2m
is it dy/dt or dy/dm?
2m e^mt,
e^mt if differentiate = e^mt or me^mt?
 
  • #11
I can't read what you are writing here, is it supposed to be y = \frac{2me^{mt}}{cos(2m)}?

Is m a constant or function of t?

As for your 2nd question, it is true that \frac{d}{dt} \ e^{mt} = me^{mt}.
 
  • #12
it is correct except the cos have no bracket, cos2m
i have no idea,im having trouble to decide whether it is dy/dm or dy/dt
 
  • #13
Is m a function of t? Secondly what are you differentiating with respect to?
 
  • #14
here is my answer,do tell whether it is right or wrong.
2e^mt(m cos 2m+cos 2m+2m sin 2m)/(cos 2m)^2
 
  • #15
just assume it as dy/dt
sorry I am not good in english.
im using quotient rule to solve that.
 
  • #16
You haven't clarified this question: Is 'm' a constant? Or is it a function of t?
 

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