It should read 1/ [(3x+1)/2]^2 + 1

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- Apr 1st 2008, 05:34 PMh2osprey
- Apr 1st 2008, 05:36 PMJhevon
- Apr 1st 2008, 05:55 PMi_zz_y_illuhuh
k look at this example, i shud be able to do this without substitution....

using (INT) 1/(a^2 + (f(x))^2 ) = 1/a arctan x/a

i can say that (int) 4/(x^2 -2x + 3)dx = 4(int) 1/(x-1)^2 +2 dx

there = 4/root2 arctan (x-1)/root2 + c

which is correct,, are you saying thatbecause with my previous example the

f(x) has a coefficient if you like of three before it i have to use substitution? - Apr 1st 2008, 05:57 PMKrizalid
See my signature for LaTeX typesetting, really, see it to make your posts clearer.

- Apr 1st 2008, 06:08 PMh2osprey
Nope, you don't have to use substitution if you can remember to deal with the coefficient later on (i.e. divide by 3). Jhevon did a great job illustrating the basic principles behind his solution, and it's very comprehensive - but ultimately you don't have to use substitution if you know how to solve it otherwise.

- Apr 1st 2008, 06:11 PMi_zz_y_illI Totally Gottit Triumph
(Bow)(Bow)(Bow)(Handshake) U HAVE TO SUBSTITUTE U=3X+1

DX=DU/3 CANCELS + GIVES ANSWER PRESUMABLY I MUST SUBSTITUTE WHENEVER THERE A COEFFICIENT OF X IN THOSE BRACKETS!? - Apr 1st 2008, 06:24 PMh2osprey
Effectively, you could put it that way, yes, or you could just do it in your head =)