Well in the first method, you have left off 1/2 until after the "v" substitution. But I expect that might just have been cut off...
But then you left off the negative...
But then that's back later...
Go back and fix up all your sloppy errors.
In the second, what your classmate did is valid. Surely you can see that $\displaystyle \begin{align*} \mathrm{e}^{2x} + 4 - \mathrm{e}^{2x} = 4 \end{align*}$, so 1/4 of that is 1, as required.
oh my gosh I'm sorry the paper wasn't scanned properly.... Here this is the right one now can you look at this?
Thanks for explaining to me how what my classmate did is valid.. wow I never would've thought of that way of transforming it.. would you recommend his method over mine?
I just checked with my calculator that my answer is equal to his (I substituted a value for x) but I don't know how to transform his answer to look like my answer and viceversa
Where you ask "How is this possible" you have, first, $\displaystyle \int \frac{dx}{e^{2x}+ 4}$ which your friend has changed to $\displaystyle \frac{1}{4}\int \frac{e^{2x}+ 4- e^{2x}}{e^{2x}+ 4}$. He has, first multiplied the numerator and denominator by 4: $\displaystyle \int\frac{4dx}{4(e^{2x}+ 4)}$ and then took the constant, 4, in the denominator out of the integral: $\displaystyle \frac{1}{4}\int\frac{4dx}{e^{2x}+ 4}$. He then added and subtracted $\displaystyle e^{2x}$ in the denominator: $\displaystyle \frac{1}{4}\frac{(e^{2x}+ 4- e^{2x})dx}{e^{2x}+ 4}$.
I presume that he did that in order to have $\displaystyle e^{2x}+ 4$ in both numerator and denominator so that he can divide: $\displaystyle \frac{e^{2x}+ 4- e^{2x}}{e^{2x}+ 4}= 1- \frac{e^{2x}}{e^{2x}+ 4}$ because, of course, $\displaystyle \frac{e^{2x}+ 4}{e^{2x}+ 4}= 1$.
Now, the integral is $\displaystyle \frac{1}{4}\int (1+ \frac{e^{2x}}{e^{2x}+ 4})dx= \frac{1}{4}\int dx+ \frac{1}{4}\int \frac{e^{2x}dx}{e^{2x}+ 1}$. The first integral is, of course, [tex]\frac{1}{4}\int dx= \frac{1}{4}x[tex] (ignoring the constant of integration for now). To do the second integral your friend let $\displaystyle u= e^{2x}+ 1$ so that $\displaystyle du= 2e^{2x}dx$ or $\displaystyle \frac{1}{2}du= e^{2x}dx$. The second integral is then just $\displaystyle \frac{1}{4}\left(\frac{1}{2}\int \frac{du}{u}= \frac{1}{8}ln|u|= \frac{1}{8}ln(e^{2x}+ 1)$ ([tex]e^{2x}+ 1[tex] is never negative so you can drop the absolute value).
I won't go through your solution but point out that both solutions are correct! You have $\displaystyle -\frac{1}{8}ln(1+ 4e^{-2x})+ C$. That is the same as $\displaystyle -\frac{1}{8}ln(e^{-2x}(e^{2x}+ 1))+ C= -\frac{1}{8}\left(ln(-2x)+ ln(e^{2x}+ 1)\right)= -\frac{1}{8}\left(-2x+ ln(e^{2x}+ 1)\right)$$\displaystyle = \frac{1}{4}x- \frac{1}{8}ln(e^{2x}+ 1)+ C$
So both of you are correct. Very good! Although I think both solutions are more complicated than necessary. I would have let $\displaystyle u= e^{2x}+ 4$ right at the start. Then $\displaystyle du= 2e^{2x}dx= 2(u- 4)du$ so $\displaystyle dx= \frac{du}{2(u- 4)}$ and the integral become $\displaystyle \frac{1}{2}\int \frac{du}{u(u- 4)}$. Using "partial fractions" we can write that as $\displaystyle \frac{1}{8}\int \frac{du}{u-4}- \frac{1}{8}\int\frac{du}{u}= \frac{1}{8}\left(ln|u- 4|- ln|u|\right)+ C= \frac{1}{8}\left(ln|e^{2x}|- ln|e^{2x}+ 4|\right)+ C= \frac{1}{8}\left(2x- ln(e^{2x}+ 4)\right)+ C$ as before.
thanks for showing another interesting efficient solution to the problem HallsofIvy! One part I don't understand is how did $\displaystyle -\frac{1}{8}ln(1+ 4e^{-2x})+ C$ become
$\displaystyle -\frac{1}{8}ln(e^{-2x}(e^{2x}+ 1))+ C$ could you explain it?
Also Jester I really really like the simplicity of your approach.
Halls you said that
$\displaystyle -\frac{1}{8}ln(1+ 4e^{-2x})+ C$ becomes
$\displaystyle -\frac{1}{8}ln(e^{-2x}(e^{2x}+ 1))+ C$
I didn't get the same thing:
$\displaystyle (1+ 4e^{-2x}) = \frac{1}{1} + \frac{4}{e^{2x}} = \frac{e^{2x}+4}{e^{2x}} = e^{-2x}(e^{2x}+4)$
by the way i just figured out how to use LaTex