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# P3 Integration Challenge watch

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1. (Original post by imasillynarb)
Which answer is right?!
I don't know, I'm the dummie asking the questions.
I know how to do these now, I forgot all about integration by partial fractions!!
Bring on the 50/100 in P3!!!
Got 52 in my mock, I need 50 ums plus 70 something in S2 to get a B overall!!!
2. (Original post by kikzen)
1,3
What the hell have you got there, it's like font 0.9.
3. (Original post by Nylex)
Heh, how'd you know about 1/(1 + x^2) being d/dx (arctan x)? If it's Edexcel P3, they wouldn't want that.. wouldn't they want partial fractions or something instead?
formula book tbh.
4. (Original post by kikzen)
formula book tbh.
Oh seen. It's easy to prove d/dx (arctan x) = 1/(1 + x^2) though.
5. (Original post by BigJohn)
What the hell have you got there, it's like font 0.9.
you can read it

(plus i would have to type it up again, and i cant be arsed with that)
6. Ill show you the way I did it, tell me where Im going wrong:

1/(xlnx) = 1/x.1/lnx

u = 1/lnx

To differentiate:

z = 1
dz/dx = 0
s= lnx
ds/dx = 1/x

[lnx x 0 - 1/x x 1]/(lnx)^2 = -1/[x(lnx)^2]

du/dx = -1/[x(lnx)^2]

dv/dx = 1/x
v = lnx

By parts uv - {vdu/dx

= 1/lnx. lnx - { -1lnx/x(lnx)^2

= 1 + {1/xlnx

2{f(x) = 1
{f(x) = 1/2

??
7. (Original post by BigJohn)
I don't know, I'm the dummie asking the questions.
I know how to do these now, I forgot all about integration by partial fractions!!
Bring on the 50/100 in P3!!!
Got 52 in my mock, I need 50 ums plus 70 something in S2 to get a B overall!!!
i=int of (x+1)^(2e^x) dx
Let u=x+1, so that du=dx therefore, I=
u^(2e^(u+1)) dx
u = e^lnu, so that I=int of e^2lnu(e^(u+1)) dx
derivative of 2lnue^(u+1)=((2/u)+(2lnu))(e^u+1)
So I=e^2lnu(e^(u+1)) / (2/u)+(2lnu))(e^u+1)
8. (Original post by imasillynarb)
Ill show you the way I did it, tell me where Im going wrong:

1/(xlnx) = 1/x.1/lnx
If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
9. Actually now I get 0

I dont think question 2 can be done.
10. (Original post by Nylex)
If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
Ahh yeh got it, surprisingly easy, I dont see why it cant be done using parts though?
11. (Original post by imasillynarb)
Ahh yeh got it, surprisingly easy, I dont see why it cant be done using parts though?
why didn't anybody listen to me??
12. (Original post by mockel)
why didn't anybody listen to me??
You didnt say how you did it! I just tried to see if I got the same answer as you
13. (Original post by Nylex)
If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
what
i never learnt that :/ i do everything raw.. but it doesnt work.. :/
14. (Original post by Nylex)
If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
a proper solution anyone ?

not just a shortcut, like.
15. (Original post by kikzen)
a proper solution anyone ?

not just a shortcut, like.
INT 1/(xlnx) dx

u = lnx
du/dx = 1/x
dx = x.du

= INT [1/(xlnx)].x.du

'x' cancels, to give: INT (1/lnx) du
But, lnx = u, so INT (1/u) du
= ln u + C
= ln (lnx) + C
16. (Original post by kikzen)
a proper solution anyone ?

not just a shortcut, like.
That's a proper solution damn it.
17. (Original post by mockel)
INT 1/(xlnx) dx

u = lnx
du/dx = 1/x
dx = x.du

= INT [1/(xlnx)].x.du

'x' cancels, to give: INT (1/lnx) du
But, lnx = u, so INT (1/u) du
= ln u + C
= ln (lnx) + C
If you're working in R0+ (the set of all strictly positive real numbers; greatest set on which ln is defined), you need to take ln¦lnx¦ + C, as lnx is negative if x<1. If lnx<0, then lnx(lnx) does not exist.
18. (Original post by mockel)
INT 1/(xlnx) dx

u = lnx
du/dx = 1/x
dx = x.du

= INT [1/(xlnx)].x.du

'x' cancels, to give: INT (1/lnx) du
But, lnx = u, so INT (1/u) du
= ln u + C
= ln (lnx) + C
genious!

ive typed it up (in a bigger font for all you geriatrics) so its a bit clearer. thanks again dude! and nick, of course
19. (Original post by kikzen)
genious!

ive typed it up (in a bigger font for all you geriatrics) so its a bit clearer. thanks again dude! and nick, of course
What program do you do that with?
20. x/(4-x^2)

x^3+4x^2/(x^2+x-2)

sin^3thetacos^2theta

x^3/(X^3-x)

Won't tell you my answers in case they're wrong.
Bloody hell, they ain't this hadr in the book.

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