| excelblue | let y = x + Sqrt[x^2 + 2x], find the limit on ln y and then after doing that, get the proper thing by doing e^(limit of ln y) again |
| JabberWalkie | probably multiply by x-Sqrt[x^2+2x] and divide.. |
| godling | that gives me -2x in the numerator hang on yeah, -2x/x-Sqrt[x^2+2x] |
| excelblue | yay, L'Hospital :) |
| kmh | zbrown : actually i kinda told you nonsense go with excelblue's substitution |
| excelblue | the place you need to go when you have a math problem :P |
| zbrown | ; lol ok |
| excelblue | well, you could use the Euler formula definitions of sin and cos, work from there using partial fractions |
| kmh | zbrown : because partial fraction of that will yield you the expression you already have |
| excelblue | just don't expect to be done within 5 pages (of tiny work, that is) |
| kmh | zbrown : if you think about it for a minute you'll see why |
| zbrown | ya just did |
| godling | excelblue: was the bit about e^... towards me? |
| excelblue | yeah |
| godling | i see |
| excelblue | it's one of the hardest parts about first year calculus IMO unless you make lots of simple arithmetic errors - then it becomes the riemann sum w/o a calculator |
| zbrown | excelblue: ok so in the case of the int((x+1)/(x^2+4)^2), one side is x, one side is 2, and one side is (x^2+4) right? |
| godling | excelblue: how is e involved in computing the limit of this function? |
| zbrown | for trig substitution that is |
| kmh | zbrown : not quite |
| zbrown | kmh: trying to figure it out, not sure where im going wrong? |
| kmh | int((x+1)/(x^2+4)^2)= int((x)/(x^2+4)^2)+ int(1/(x^2+4)^2) first part is handled with u=x^2+4, second part with x=tan(u) |
| zbrown | kmh: well ok... so for the sides of the triangle, one side is x, one is 2, and one is x^2+4? |
| kmh | excelblue*s suggestion was concerning the 2nd part of the sum zbrown : oh you mean a triangle to visualize the trig substitution ? |
| zbrown | kmh: ya |
| kmh | ok zbrown : i don't think you need that but yes it looks ok zbrown : basically it is pythagoras |
| zbrown | kmh: ok thanks kmh: just helps me visualize w/ the triangle :) |
| action | zbrown likes pictures |
| zbrown | likes pictures |