Edexcel C3,C4 June 2013 Thread

You can do it either way.

No. You can only use the ln rule when the top of the fraction is the differential of the bottom, or a scalar multiple of it. u is not a scalar as it is not a number, it is a function. To integrate that you'd need to use long division or the methods mentioned earlier to separate it into stuff you can integrate

You don't need to divide through for equal powers of x.

The only difference here would be that, instead of splitting it into:

$\dfrac{A}{2x+1} + \dfrac{B}{2x-1}$

You'd split it into:

$\dfrac{A}{2x+1} + \dfrac{B}{2x-1} + \dfrac{C}{(2x+1)(2x-1)}$

C4 Solomon Worksheets: http://www.london-oratory.org/maths/Solomon%20C4%20Worksheets/C4Edex/sections.htm

Oh right.. So you wouldnt be able to split it like u/4 + u/u ?

what paper is this from?

If it was (u+4)/u then you could split it into u/u + 4/u, but you can't split the denominator up like that - think when you add fractions, you can't add them unless the denominator is the same

Good luck in the C4 exam tomorrow guys, I hope you all have been revising S2

How do you integrate 2^x

Thank you! I still dont understand how you would split u/u+4 I tried doing (u+4) (something) = U ....

Is the something -4?...How would you form it into a fraction then?

I'm assuming you know that $y = 2^x \ \Rightarrow \ \dfrac{dy}{dx} = 2^x \ln 2$

So,

$\displaystyle \int 2^x \ dx$

$= \displaystyle \frac{1}{\ln 2} \int 2^x \ln 2 \ dx$

$= \dfrac{2^x}{ \ln 2} + C$

Thank you! I still dont understand how you would split u/u+4 I tried doing (u+4) (something) = U ....

Is the something -4?...How would you form it into a fraction then?

By the way this is for the question integrate 1/4+root x+1 .. where u = root x+1.

I just get confused a little with some substitution q's.

I'm assuming you know that $y = 2^x \ \Rightarrow \ \dfrac{dy}{dx} = 2^x \ln 2$

So,

$\displaystyle \int 2^x \ dx$

$= \displaystyle \frac{1}{\ln 2} \int 2^x \ln 2 \ dx$

$= \dfrac{2^x}{ \ln 2} + C$

U/(U+4) = (U-4+4)/(U+4) does this help now?