Question

Integration by substitution

Answer 1

I believe you've written your integral in invisible ink

Answer 2

\[\int\limits_{}^{}1+ e ^{3x-2} \div e ^{3x} +3x\]

Answer 3

may you use residue theorem?

Answer 4

whats that ?

Answer 5

Your division symbol is making this a little tough to read, can you clarify what the problem is suppose to look like?\[\large \int\limits\frac{1+e^{3x-2}}{3x+e^{3x}}dx\]Or\[\large \int\limits \frac{1+e^{3x-2}}{e^{3x}}+3x \;dx\]

Answer 6

the first one

Answer 7

but its not 3x-2 in the numerator its just e^3x

Answer 8

Oh lol, that makes this quite a bit easier then :)

Answer 9

\[\large \int\limits \frac{1+e^{3x}}{3x+e^{3x}}dx\]
Make a `u substitution` letting \(\large u=3x+e^{3x}\).

Answer 10

Taking the derivative of our substitution,\[\large du=(3+3e^{3x}) \;dx\]Factoring a 3 out of each term,\[\large du=3(1+e^{3x}) \;dx\]Divide both sides by 3,\[\large \frac{1}{3}du=(1+e^{3x}) \;dx\]

Answer 11

Understand how to plug in your substitution? :)

Answer 12

i'l let you know let me think
thanks tho

Answer 13

but my teacher told me everythig hasw to be i the same term or letter , so there cant be a x & a u in the equation

Answer 14

ignore that last comment but bboth sides equaled each other out at the end is that normal ?

Answer 15

what do you mean? when you u substitute what you get for du hould cancel all the x terms

Answer 16

so as @zepdrix said i plugged in du for du but what do i do next ?

Answer 17

You'll be replacing all of the \(x\)'s and also \(dx\) with something involving \(u\)'s and \(du\) as Outkast mentioned.

Answer 18

solve for dx and plug that in

Answer 19

\[du=3(1+e^{3x})dx\]
\[\frac{du}{3(1+e^{3x})}=dx\]

Answer 20

putting this in there and also \[u=3x+e^{3x}\], everything will be in u and the x terms will cancel

Answer 21

So we're assigning this substitution,\[\large \color{orangered}{u=3x+e^{3x}}\]\[\large \color{royalblue}{\frac{1}{3}du=(1+e^{3x}) \;dx}\]

\[\large \int\limits\limits \frac{\color{royalblue}{1+e^{3x}}}{\color{orangered}{3x+e^{3x}}}\color{royalblue}{dx} \qquad \rightarrow \qquad \int\limits\limits \frac{\color{royalblue}{\frac{1}{3}du}}{\color{orangered}{u}}\]

Answer 22

why did you divide ??

Answer 23

?

Answer 24

?

Answer 25

Divide what? What are you talking about?

Answer 26

why did you divide du by u

Answer 27

It might be easier to read the integral if we write it like this,\[\large \int\limits\limits \frac{1+e^{3x}}{3x+e^{3x}}dx \qquad \rightarrow \qquad \int\limits\limits \frac{1}{3x+e^{3x}}(1+e^{3x})dx\]

Answer 28

\[\large \int\limits\limits\limits \frac{1}{\color{orangered}{3x+e^{3x}}}\color{royalblue}{(1+e^{3x})dx} \qquad \rightarrow \qquad \int\limits\limits\limits \frac{1}{\color{orangered}{u}}\color{royalblue}{\left(\frac{1}{3}du\right)}\]

Answer 29

You were confused because the du was on top I guess? :o

Answer 30

is that the final answer

Answer 31

No, we haven't even integrated yet.
We've simplified the integral so it will be a lot easier to integrate from here.

Answer 32

ooooooooooh

Answer 33

You can pull the 1/3 outside of the integral since it's just a constant.
\[\large \frac{1}{3}\int\limits \frac{1}{u}du\]

This is a good one to have memorized. Do you remember this integral? :)

Answer 34

ya

Answer 35

whats the antiderivative of 3x ?

Answer 36

\[\large \frac{3x^2}{2}\]
Hmm I'm not sure why that would apply here though :o

Answer 37

to find the integral of e^3x