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C3: How to integrate?

Afternoon all! :smile:

I know how to integrate simple functions like 3x^2 (add one to the power divide by the new power) but I can't do harder ones.
I'm stuck on integrating: (2x+1)/(x) and (2x+1)^2/(x)^2.
Not got a clue where to start :s-smilie:

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Reply 1

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

(a+b)/c = a/c + b/c

Reply 2

jim666666

Original post by Get me off the £\?%!^@ computer

(a+b)/c = a/c + b/c

cheers, i got that bit, its the actual integrating i can't do.

Reply 3

anshul95

Original post by jim666666

cheers, i got that bit, its the actual integrating i can't do.

If you mean you are stuck for the integral of 1/x it is ln x. For the second one you could in this case just expand (2x+1)^2 and divide by x^2 and then integrate.

Reply 4

Dark-Myth

Either expand, or treat it as two functions:
f(x) = uv , where u=2x+1, and v=2x^-1

Then f'(x)=u*dv/dx + v*du/dx

Reply 5

luuucyx

chain rule? or am i just being stupid saying that? what board are you on?

Reply 6

anshul95

Original post by luuucyx

chain rule? or am i just being stupid saying that? what board are you on?

Chain rule is for differentiation we want to integrate the functions given

Reply 7

luuucyx

Original post by anshul95

Chain rule is for differentiation we want to integrate the functions given

huumm i knew that :confused:

haha
substitution?
all ive been shown are realllyyy simple ones of these...

Reply 8

Pheylan

Original post by Dark-Myth

Either expand, or treat it as two functions:
f(x) = uv , where u=2x+1, and v=2x^-1

Then f'(x)=u*dv/dx + v*du/dx

i don't even what

Reply 9

Dark-Myth

Original post by Pheylan

i don't even what

I know it's a bit late...
but, it's the chain rule.

Reply 10

lemmy123

If it's divide, I'd use the quotient rule?

Reply 11

Pheylan

Original post by Dark-Myth

I know it's a bit late...
but, it's the chain rule.

actually it's incorrect use of the product rule in a situation where the product rule isn't actually supposed to be used

Reply 12

Dark-Myth

Original post by Pheylan

actually it's incorrect use of the product rule in a situation where the product rule isn't actually supposed to be used

OH ****. It's integration. I didn't realise... my baaaad.
We don't do integration in C3, so I thought it was differentiation. :frown:

(edited 13 years ago)

Reply 13

Eloades11

Original post by luuucyx

huumm i knew that :confused:

haha
substitution?
all ive been shown are realllyyy simple ones of these...

substution is C4 :smile:

nice try though

Reply 14

Faith01

Original post by Get me off the £\?%!^@ computer

(a+b)/c = a/c + b/c

I am confused? There is no integration in C3 :s-smilie:

Reply 15

mir3a

Original post by luuucyx

chain rule? or am i just being stupid saying that? what board are you on?

LOL

Reply 16

mir3a

Original post by Faith01

I am confused? There is no integration in C3 :s-smilie:

He may be doing a different exam board to what your doing..

Reply 17

Faith01

Original post by mir3a

He may be doing a different exam board to what your doing..

Oh yeah that makes sense!

Reply 18

mir3a

Original post by jim666666

Afternoon all! :smile:

for the first one do.. let f(x)=2x+1/x, then simplify the expression dividing both 2x and 1 by x. so, f(x)=2+x^-1 then integrate as you would normally.

For the second one, let f(x)=(2x+1)^2/x^2, expand the numerator, divide by the denominator as we did above, and integrate :biggrin:

Reply 19

sohail.s

Original post by mir3a

for the first one do.. let f(x)=2x+1/x, then simplify the expression dividing both 2x and 1 by x. so, f(x)=2+x^-1 then integrate as you would normally.

I think im missing something here, help is appreciated, if you integrate x^-1 its not possible is it? you'd end up dividing by 0 ??

Bear in mind i'm still in AS year and studying C2 atm.