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

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In summary, the conversation is about solving an integration problem and differentiating a function. The correct answer for the integration problem is 1/2 ln (1+t^2)+c. For the differentiation problem, the correct answer is 2e^mt(m cos 2m+cos 2m+2m sin 2m)/(cos 2m)^2. The conversation also includes a question about whether 'm' is a constant or a function of t.
  • #1
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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  • #2
How did you get your answer?
 
  • #3
im using the integration using substitution..
im confuse right now.. can anyone give me the full calculation or what method i should use.
 
  • #4
[tex]\frac{t}{1+t^2} dt[/tex]

Let [itex] u=1+t^2[/itex] what is [itex]\frac{du}{dt}[/itex]? then what is dt in terms of du?
 
  • #5
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:
  • #6
could anyone help me here? pls..
 
  • #7
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.
 
  • #8
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?
 
  • #9
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 [tex]y = \frac{2me^{mt}}{cos(2m)}[/tex]?

Is m a constant or function of t?

As for your 2nd question, it is true that [tex]\frac{d}{dt} \ e^{mt} = me^{mt}[/tex].
 
  • #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?
 

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

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value of a quantity that is changing continuously.

2. How do you solve an integral?

To solve an integral, you need to use integration techniques such as substitution, integration by parts, or trigonometric substitution. You also need to have a good understanding of the fundamental theorem of calculus and the properties of integrals.

3. What is the purpose of the constant "c" in the solution?

The constant "c" is known as the constant of integration and is added to the solution of an indefinite integral to account for all possible solutions. It represents the unknown value that would make the solution valid for any value of the variable.

4. Can you use a calculator to solve an integral?

Yes, there are calculators that can solve definite and indefinite integrals. However, it is important to also understand the concepts and techniques behind integration to ensure the accuracy of the solutions.

5. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and represents the area under a curve between those limits. An indefinite integral does not have limits and represents the general solution of an integral, with the added constant "c".

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