Question

any strategy suggestions for dealing with integration dealing with "e"

Answer 1

examples\[\int\limits\frac{ 5 }{ 3e^x -2 }\]

Answer 2

multiply and divide by e^(-x)
put e^(-x) = u

Answer 3

or \[\int\limits \frac{ e^(1/t) }{ t^2 }dt\]

Answer 4

put 1/t = u
-1/t^2 dt =du

Answer 5

so when dealing with these what is the best approach? My mind draws a blank on these for some reason. I think things like multiplying by the conjugate and some other things that don't seem to work out

Answer 6

For 1st one, remember that int f'(x)/f(x) dx = ln |x| +c
so wee need derivative of denominator in numerator,
since derivative of exponential function is exponential , we can multiply and divide by exponential to get derivative in numerator.
For 2nd, just try to make the power of t as linear.

Answer 7

I have come to the conclusion that I am at the wrong school. I need to attend hogwarts because calc II is 90% MAGIC!!!!!!!! lol

Answer 8

lol!
but did u understand why and what u need to substitute ?

Answer 9

The first example I just worked out and it made sense, I am about to follow through the second. I really hope I can start seeing these things on my own

Answer 10

*For 2nd, just try to make the power of e as linear

Answer 11

Ok I thought I had the first but...... I got to a point where i had \[-5\int\limits \frac{ 1 }{ 3-2u }du\] what do I do with that? I can' see a way to make that \[\frac{ du }{ u }\] so that I can use ln

Answer 12

put t=3-2u now (its easier)
OR you could directly put 3-2e^-x = u

Answer 13

where did "t" come from? that was the second one I think

Answer 14

't' is just another variable, u could use 'y' or 'chrisplusian' instead of 't'
its like 2nd substitution
or remember that
\(\large \int \frac{1}{ax+b}dx= (1/a)ln|ax+b|+c \)

Answer 15

so then you are saying that t=3-2u and dt=-2du?

Answer 16

yup

Answer 17

let me try to wrap my mind around that for a minute while I work it out on paper

Answer 18

ok so then I end up with \[-\frac{ 5 }{ 2 }\ln \left| 3-2u \right|+c\] which then becomes \[-\frac{ 5 }{ 2 }\ln \left| 3-2e ^{-x} \right|+c\] is that what you get? People are supposed to figure this out on their own? AGHHHHHHH!!!!!

Answer 19

Thanks @hartnn

Answer 20

yup, thats correct.