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C4 integration help

Can you in integrate y=1/(2x-1)^1/2 without using substitution?

If so can you help me please?
Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by Vorsah
Can you in integrate y=1/(2x-1)^1/2 without using substitution?

If so can you help me please?


Have you learnt how to integrate "by inspection" or "reverse chain rule"

The key is to look at your bracket and raise the power by 1 then ask yourself what the differential is



So, what would you get if you differentiated (2x−1)12(2x-1)^{\frac{1}{2}}


If you know that then you can adapt the answer to solve your original problem
Yeah just think about the process in reverse, i.e take an educated guess at the anti derivative


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Indeed, what he said


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Reply 4
Avatar for Vorsah
Vorsah
OP
5
Original post by TenOfThem
Have you learnt how to integrate "by inspection" or "reverse chain rule"

The key is to look at your bracket and raise the power by 1 then ask yourself what the differential is



So, what would you get if you differentiated (2x−1)12(2x-1)^{\frac{1}{2}}


If you know that then you can adapt the answer to solve your original problem


I got (2x-1)^1/2 + c using reverse chain rule

Thanks
Reply 5
Avatar for Vorsah
Vorsah
OP
5
Original post by TenOfThem
Have you learnt how to integrate "by inspection" or "reverse chain rule"

The key is to look at your bracket and raise the power by 1 then ask yourself what the differential is



So, what would you get if you differentiated (2x−1)12(2x-1)^{\frac{1}{2}}


If you know that then you can adapt the answer to solve your original problem


How would I integrate 3e^x(e^x+2)^-1 using recognition.

When you differentiate the brackets you get e^x which is a factor of 3e^x, so integration by recognition can be used

But after that I'm confused because of the ^-1, usually you would have to add 1 to the power then differentiate.
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by Vorsah
How would I integrate 3e^x(e^x+2)^-1 using recognition.

When you differentiate the brackets you get e^x which is a factor of 3e^x, so integration by recognition can be used

But after that I'm confused because of the ^-1, usually you would have to add 1 to the power then differentiate.


When you have a power of -1 then you are looking at ln

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