took me to a world of confusion :( sheesh what am I missing?
also, can one use the standard integral with the second fraction and get arc cosh [(x+2)/3] ?
well before completing the square it seemed quite clear to me that letting u= x2+ 4x -5 would make 1/2 du = x+2. for solving the 1st fraction.. but looking at the completed square i assume i should be doing something else?
that gives me a denominator of (x+2)2 - 9 ... why do we do this?
also: can I split the initial fraction up into 2 separate ones? IE. (x+2) and (+1) (both over same denominator) in this way I can use substitution for the 1st fraction right?
PS. sorry about the duplicate post. Realized it was...
Homework Statement
Integrate: (x + 3) / sq rt of (x2 + 4x - 5) Homework Equations
The Attempt at a Solution
No idea. they are not derivative / integral of each other so I am not sure :( Please help asap. Thanks in advance
Is this applied to the box at rest? in that case the applied force has to be greater than the max Static Friction for the box to have any acceleration.
Hi farmguy
youre using the right approach but you've left out information and made a calculation error:
if E0=Ef ...Initial total mechanical energy = final tot mech enrgy
then...
remember that if u make final height 0, like u did, u have to make initial height 2m. From reading the...
so simple :D dammit how do u know when to use these principles of conversation? its so confusing :(
so by doing that i got v= -29,4 but i need angular speed.
so VT = \omega r ?
then i get -19,6 rad/s