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Can someone check this C4 type integral for me? I feel like the answers got it wrong

CancerousProblem

\int^1_0 \frac{\ln (\frac{x^2+3x+2}{(x+3)^{2}})}{(x+3)^{2}}\ dx

edit: nevermind i see what i did wrong now

(edited 8 years ago)

Original post by CancerousProblem

Unparseable latex formula:

[br][font="verdana"][color="#444444"]\int^1_0 [/color][/font][font="verdana"][color="#444444"]\frac{[/color][/font][font="verdana"][color="#444444"]\ln [/color][/font][font="verdana"][color="#444444"]\frac{x+1}{x+3}[/color][/font][font="verdana"][color="#444444"]}{(x+3)^{2}}[/color][/font][font="verdana"][color="#444444"]\[/color][/font][br]

\int^1_0 \frac{\ln \frac{x+1}{x+3}}{(x+3)^{2}}\ dx

I got 19/12ln3 - 7/3ln2 - 1/6
Answers got -1/12ln2 less than me

We can't see the image.

rage inducing member?

CancerousProblem

OP

Original post by lizard54142

We can't see the image.

never added an image, my latex just got bugged

Original post by CancerousProblem

never added an image, my latex just got bugged

- Latex bugged
- wrong question
- wrote out your working wrong
- incomprehensible substitution

I am completely lost!

CancerousProblem

OP

Original post by lizard54142

- Latex bugged
- wrong question
- wrote out your working wrong
- incomprehensible substitution

I am completely lost!

lol **** i am so sorry

the question is correct now
x+1/x+3 = x was the substitution I used to get the integral

Original post by CancerousProblem

lol **** i am so sorry

the question is correct now
x+1/x+3 = x was the substitution I used to get the integral

Haha no problem, so:

u = x + \dfrac{1}{x+3}

is the substitution?
Or:

x = u + \dfrac{1}{u+3}

?

CancerousProblem

OP

Original post by lizard54142

Haha no problem, so:

u = x + \dfrac{1}{x+3}

is the substitution?
Or:

x = u + \dfrac{1}{u+3}

?

um no
(x+1)/(x+3) = u

CancerousProblem

OP

Original post by lizard54142

Haha no problem, so:

u = x + \dfrac{1}{x+3}

is the substitution?
Or:

x = u + \dfrac{1}{u+3}

?

nevermind I see where I messed up now thanks. its been a fun 30 minutes with latex rofl
I should just take a picture next time

Original post by CancerousProblem

nevermind I see where I messed up now thanks. its been a fun 30 minutes with latex rofl
I should just take a picture next time

Gosh I was close to doing it as well... where did you go wrong?

CancerousProblem

OP

Original post by lizard54142

Gosh I was close to doing it as well... where did you go wrong?

I subsituted (x+2)/(x+3) as (x+1)/(x+3) + 1 rofl

I thought the answers had the same thing but that was before that step

Original post by CancerousProblem

I subsituted (x+2)/(x+3) as (x+1)/(x+3) + 1 rofl

I thought the answers had the same thing but that was before that step

That's where I got up to when you said you'd done it haha:

ln(\dfrac{x+2}{x+3}) = ln(\dfrac{1}{x+3} + \dfrac{x+1}{x+3})

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