integration help (1 Viewer)

[100percent]

how do you integrate
2x - 3
-------------- dx
x² - 2x + 2

with boundaries 2 & 1. thanks.

btw question is from cambridge page 155 question 17

 

[shafqat]

how do you integrate
2x - 3
-------------- dx
x² - 2x + 2

with boundaries 2 & 1. thanks.

btw question is from cambridge page 155 question 17

= 2x-2 / x² - 2x + 2 - 1/x² - 2x + 2
= 2x-2 / x² - 2x + 2 - 1/(x-1)^2 + 1
= log (x² - 2x + 2) - tan^-1(x-1)

edit: from 1 to 2, ah couldnt be bothered, i think you can do it yourself

 

[100percent]

it never occured to me to split the 2x - 3 to 2x - 2 -1.
thanks man

 

[..:''ooo]

100percent i think u need a treatment of corroneous integration questions

 

[KFunk]

I've been doing that coroneous revision set of 100 questions. It's a hell of a lot of integration.

 

[hasterz]

hello people ive got 1 i need help

integral of (2^x)^(1/2)

 

[yrtherenonames]

uh from memory, I think u can change the (2^x)^1/2 to 2^(x/2) and then change the 2 to a e^ln2.
So u now have integral of e^((0.5x)ln2)dx which with a bit of trickery you can do.
Could be wrong but I think that's how you do it.

 

[KFunk]

I think that's pretty much it.

∫ 2^x/2 dx

convert it to base e to make your life less difficult ---->
e^u = 2
log_e2=u
hence 2 = e^ln2

∫ 2^x/2 dx
= ∫ (e^ln2)^x/2 dx
= ∫ e^(x.ln2)/2 dx
= (2/ln2) ∫ ln2/2.e^(x.ln2)/2 dx
= (2/ln2).e^(x.ln2)/2 + C
= (2/ln2)√(2^x) + C

 

[yrtherenonames]

Looks good to me