- #1

rojak44

- 13

- 0

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..

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- Thread starter rojak44
- Start date

- #1

rojak44

- 13

- 0

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..

- #2

exk

- 119

- 0

How did you get your answer?

- #3

rojak44

- 13

- 0

im confuse right now.. can anyone give me the full calculation or what method i should use.

- #4

rock.freak667

Homework Helper

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Let [itex] u=1+t^2[/itex] what is [itex]\frac{du}{dt}[/itex]? then what is dt in terms of du?

- #5

rojak44

- 13

- 0

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)

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

rojak44

- 13

- 0

could anyone help me here? pls..

- #7

Defennder

Homework Helper

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Cancel out t in this step.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)

- #8

rojak44

- 13

- 0

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

Defennder

Homework Helper

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Yes, that is correct.

- #10

rojak44

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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

Defennder

Homework Helper

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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

rojak44

- 13

- 0

i have no idea,im having trouble to decide whether it is dy/dm or dy/dt

- #13

Defennder

Homework Helper

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Is m a function of t? Secondly what are you differentiating with respect to?

- #14

rojak44

- 13

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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

2e^mt(m cos 2m+cos 2m+2m sin 2m)/(cos 2m)^2

- #15

rojak44

- 13

- 0

just assume it as dy/dt

sorry I am not good in english.

im using quotient rule to solve that.

sorry I am not good in english.

im using quotient rule to solve that.

- #16

Defennder

Homework Helper

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You haven't clarified this question: Is 'm' a constant? Or is it a function of t?

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