C3: How to integrate?

huumm i knew that haha
substitution?
all ive been shown are realllyyy simple ones of these...

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.

sorry my explanation wasn't as clear...integrating 1/x gives you ln x, its a rule u should learn.
Let me know if you get stuck

substution is C4 nice try though

You could use chain rule, but backwards. That would work!!

1/a x 1/(n+1) x (ax+b)^(n+1)

nah ocr its C3...

really? which OCR, im on OCR and I have my C3 exam next month, ive been through 6 past papers and never come across any integration by substitution.

Unless its OCR gateway or 21 century of any other weird OCR thing, in that case ill let you off

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.

(2x + 1)/x
goes to
(2x/x) + (1/x)

then integrate

to 2x + in [x] + c



2x + in

How has this thread stretched to 2 pages...


This is equivalent to showing that the derivative of $\ln x$ is $\frac{1}{x}$ because integration is the reverse of differentiation.
Note that $\frac{d}{dx}[\ln x] = \displaystyle\lim_{h\to 0} \frac{\ln (x+h)-\ln (x)}{h} = \displaystyle\lim_{h\to 0} \frac{\ln (1+\frac{h}{x})}{h} = \displaystyle\lim_{h\to 0} \ln (1 + \frac{h}{x})^{\frac{1}{h}}$
Now if you let $\frac{h}{x}=\frac{1}{u} \implies \frac{1}{h} = \frac{u}{x}$.
Therefore we have:
$\frac{d}{dx}[\ln x]=\displaystyle\lim_{u\to \infty} \ln (1+\frac{1}{u})^{\frac{u}{x}} = \displaystyle\lim_{u\to \infty} \frac{1}{x}\ln (1+\frac{1}{u})^{u}$.

Note that $e= \displaystyle\lim_{u\to \infty}(1+\frac{1}{u})^{u}$
Therefore $\frac{d}{dx}[\ln x] = \frac{1}{x}\ln (e) = \frac{1}{x}$ as required.
Therefore $\displaystyle\int\frac{1}{x}dx = \ln x + C$.

ah ok, cheers, i hope i dont have to remember all that :P