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

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SUMMARY

The integral of the function t/(1+t^2) dt can be solved using substitution, leading to the correct answer of 1/2 ln(1+t^2) + c. The user initially proposed 1/2 ln(t(1+t^2)) + c, which is incorrect. The substitution method involves letting u = 1 + t^2, resulting in the differential dt expressed in terms of du. The discussion clarifies the correct steps and confirms the final result through collaborative problem-solving.

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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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