You are Here: Home >< Maths

# Why did he divide by 2?? :/

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016

1. Why was 2 divided by 2 & why was the same thing applied to 3?

Video:
https://youtu.be/eufqV0B6SP8?t=14m44s
Attached Images

Why was 2 divided by 2 & why was the same thing applied to 3?

Video:
https://youtu.be/eufqV0B6SP8?t=14m44s
What happens when you differentiate ln(2x+1)?
3. (Original post by Gregorius)
What happens when you differentiate ln(2x+1)?
1/(2x+1)
1/(2x+1)
Now try integrating that and see if you get the same result as when you started...
1/(2x+1)
Uh uh; how does the chain rule go?
1/(2x+1)
d/dx ln f(x) = f '(x) / f(x)
Thus d/dx ln(2x+1) = 2/2x+1
Therefore Integrating 1/2x+1 would give [1/2 ln(2x+1) ] + C

As you are multiplying by a constant of 2, the integral turns in [2/2 ln(2x+1)] +C
7. (Original post by Gregorius)
Uh uh; how does the chain rule go?
y = Int where t= 2x+1
dy/dt= 1/t

dt/dx= 2

so 1/t x 2
1/(2x+1) x 2
2/(2x+1)
8. Same reason as before you need to make the top of the fraction f '(x) = 2
as the differential of u =(2x+1) is 2
you do this by
∫ 1/(2x+1) dx
= (1/2)∫ 2/(2x+1) dx
9. (Original post by Katiee224)
Now try integrating that and see if you get the same result as when you started...
(Original post by Gregorius)
Uh uh; how does the chain rule go?
(Original post by XOR_)
Same reason as before you need to make the top of the fraction f '(x) = 2
as the differential of u =(2x+1) is 2
you do this by
∫ 1/(2x+1) dx
= (1/2)∫ 2/(2x+1) dx
Okay I get it now Thank you!!
y = Int where t= 2x+1
dy/dt= 1/t

dt/dx= 2

so 1/t x 2
1/(2x+1) x 2
2/(2x+1)
This is where the error is
you don't times by dt/dx
you times by dx/dt

so it's
∫ 1/t x (1/2) dt
= (1/2)∫ 1/t dt
= (1/2)ln(t) + c
= (1/2)ln(2x+1) + c

^ this is without the constant 2 in the original question if there was a constant 2
it would be
= (2/2)ln(2x+1) + c
11. (Original post by XOR_)
This is where the error is
you don't times by dt/dx
you times by dx/dt

so it's
∫ 1/t x (1/2) dt
= (1/2)∫ 1/t dt
= (1/2)ln(t) + c
= (1/2)ln(2x+1) + c

^ this is without the constant 2 in the original question if there was a constant 2
it would be
= (2/2)ln(2x+1) + c
But am I not supposed to use the chain rule?
But am I not supposed to use the chain rule?
if you use t as a variable then
(dy/dt) = (dy/dx)(dx/dt)

which is how you get ∫ ..... dt
so
∫ 1/t (dx/dt) dt
where dx/dt = (1/2)

-> (1/2)∫ 1/t dt

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 20, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams