Math Help - Which is the correct solution? (Integral Calc)

1. Which is the correct solution? (Integral Calc)

Here in the scanned paper I have written both my solution and my classmate's solution to the same problem
I'm wondering which solution is the correct one, since the answer looks different, and there is a part that classmate did that I don't get how he did it.. anyway it's all there on the paper

2. Re: Which is the correct solution? (Integral Calc)

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.

3. Re: Which is the correct solution? (Integral Calc)

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

4. Re: Which is the correct solution? (Integral Calc)

Where you ask "How is this possible" you have, first, $\int \frac{dx}{e^{2x}+ 4}$ which your friend has changed to $\frac{1}{4}\int \frac{e^{2x}+ 4- e^{2x}}{e^{2x}+ 4}$. He has, first multiplied the numerator and denominator by 4: $\int\frac{4dx}{4(e^{2x}+ 4)}$ and then took the constant, 4, in the denominator out of the integral: $\frac{1}{4}\int\frac{4dx}{e^{2x}+ 4}$. He then added and subtracted $e^{2x}$ in the denominator: $\frac{1}{4}\frac{(e^{2x}+ 4- e^{2x})dx}{e^{2x}+ 4}$.

I presume that he did that in order to have $e^{2x}+ 4$ in both numerator and denominator so that he can divide: $\frac{e^{2x}+ 4- e^{2x}}{e^{2x}+ 4}= 1- \frac{e^{2x}}{e^{2x}+ 4}$ because, of course, $\frac{e^{2x}+ 4}{e^{2x}+ 4}= 1$.

Now, the integral is $\frac{1}{4}\int (1+ \frac{e^{2x}}{e^{2x}+ 4})dx= \frac{1}{4}\int dx+ \frac{1}{4}\int \frac{e^{2x}dx}{e^{2x}+ 1}$. 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 $u= e^{2x}+ 1$ so that $du= 2e^{2x}dx$ or $\frac{1}{2}du= e^{2x}dx$. The second integral is then just $\frac{1}{4}\left(\frac{1}{2}\int \frac{du}{u}= \frac{1}{8}ln|u|= \frac{1}{8}ln(e^{2x}+ 1)$ ([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 $-\frac{1}{8}ln(1+ 4e^{-2x})+ C$. That is the same as $-\frac{1}{8}ln(e^{-2x}(e^{2x}+ 1))+ C= -\frac{1}{8}\left(ln(-2x)+ ln(e^{2x}+ 1)\right)= -\frac{1}{8}\left(-2x+ ln(e^{2x}+ 1)\right)$ $= \frac{1}{4}x- \frac{1}{8}ln(e^{2x}+ 1)+ C$

So both of you are correct. Very good! Although I think both solutions are more complicated than necessary. I would have let $u= e^{2x}+ 4$ right at the start. Then $du= 2e^{2x}dx= 2(u- 4)du$ so $dx= \frac{du}{2(u- 4)}$ and the integral become $\frac{1}{2}\int \frac{du}{u(u- 4)}$. Using "partial fractions" we can write that as $\frac{1}{8}\int \frac{du}{u-4}- \frac{1}{8}\int\frac{du}{u}= \frac{1}{8}\left(ln|u- 4|- ln|u|\right)+ C= \frac{1}{8}\left(ln|e^{2x}|- ln|e^{2x}+ 4|\right)+ C= \frac{1}{8}\left(2x- ln(e^{2x}+ 4)\right)+ C$ as before.

5. Re: Which is the correct solution? (Integral Calc)

Another way

$\int \dfrac{dx}{e^{2x} + 4} = \int \dfrac{e^{-2x}dx}{1+4 e^{-2x}}$

and then make the substitution $u = 1+4 e^{-2x}$.

6. Re: Which is the correct solution? (Integral Calc)

thanks for showing another interesting efficient solution to the problem HallsofIvy! One part I don't understand is how did $-\frac{1}{8}ln(1+ 4e^{-2x})+ C$ become
$-\frac{1}{8}ln(e^{-2x}(e^{2x}+ 1))+ C$ could you explain it?
Also Jester I really really like the simplicity of your approach.

7. Re: Which is the correct solution? (Integral Calc)

Why not try to manipulate the two expressions and see if you get the same thing?

8. Re: Which is the correct solution? (Integral Calc)

Originally Posted by Prove It
Why not try to manipulate the two expressions and see if you get the same thing?
Halls you said that
$-\frac{1}{8}ln(1+ 4e^{-2x})+ C$ becomes
$-\frac{1}{8}ln(e^{-2x}(e^{2x}+ 1))+ C$
I didn't get the same thing:
$(1+ 4e^{-2x}) = \frac{1}{1} + \frac{4}{e^{2x}} = \frac{e^{2x}+4}{e^{2x}} = e^{-2x}(e^{2x}+4)$
by the way i just figured out how to use LaTex

9. Re: Which is the correct solution? (Integral Calc)

Halls of Ivy made a small typo in the middle of the explanation. It should have read $-\dfrac{1}{8}\ln\left(e^{-2x}(e^{2x}+4)\right)$