Question

Hi everyone! :o) Need a little help here remember a procedure...

The problem is: integral (1/(V-IR) d(EYE)
I know the integral of 1 over something is ln(something) so I would have ln(V-IR) but the answer is actually ln(V-IR)/-R...

I can't remember what I am missing. A chain rule or something? And how to exactly do I implement that rule to get the correct answer? Thanks!

Answer 1

\(\int (1/x)dx = \ln x +c \\ but \\ \int (1/[ax+b])dx \ne \ln[ax+b]+c\)

Answer 2

the neat way to do this is to substitute the linear expression as u

u =V-IR

du = ... ?

Answer 3

lost connection...sorry...working on the du right now

Answer 4

uhggg...can't belive i forgot this!...need a hint

Answer 5

say u = ax+b

since a and b are constants

du/dx = a*1 +0 = a

so, du/dx = a


du = adx

Answer 6

and i am pretty sure its not "EYE", its I

integral (1/(V-IR) d(I)

Answer 7

i know for example if u=x^3+3x that du would equal 3x^2+3 du
i just cant seem to translate that to u=V-IR...would it be
du=(1)-(??)

Answer 8

we are differentiating w.r.t I
so V, R are constants!

u= V-IR
du/dI = 0 - R*1
makes sense ?

Answer 9

dV/dI = 0 because V is a constant

Answer 10

let me think about it for a second...

Answer 11

d/dx (ax) = a dx/dx = a

d/dI (-IR) = -R dI/dI = -R

Answer 12

i am missing something fundamental that used to make me understand this perfectly...i need to back up a second so try and follow what I am gonna ask...

Answer 13

i am gonna write out each step...just make sure i am correct

Answer 14

u=x^3+3x....
d/du = d/dx(x^3+3x)...
what would be the next step on the left? I know the right would be 3x^2+3

Answer 15

is writing d/du correct or should it be du/du? uhg

Answer 16

if i can just figure out the correct thing to write down on the left, then i know i just multiply both sides by du

Answer 17

you're differentiating w.r.t "x" right ?
so it will be du/dx on left

Answer 18

ok thank you...that is what i was missing...so one sec then...i think i can finish