Integrate 2/(e^(2x) + 4) dx using substitution

[Monkeyseat]

Question:

Using the substitution u = e^(2x) + 4, find the integral of:

2/(e^(2x) + 4) dx

Working:

I don't know how to type 'math' on these forums, so I've written out where I got up to:

I think it's correct up to the second from last line. After that I don't know how to finish it.

Please could someone guide me in the right direction?

Many thanks.

 

[fasteddie65]

Use Integration by Partial Fractional Decomposition at this point:

? du/[u(u - 4)]

1/[u(u - 4)] = A/u + B/(u - 4)
1 = A(u - 4) + Bu

If u = 0, 1 = -4A, so A = -1/4
If u = 4, 1 = 4B, so B = 1/4

? du/[u(u - 4)] = (1/4) [?du/(u - 4) - ? du/u ] = (1/4) [ ln |u - 4| - ln |u|] = (1/4) ln |(u - 4)/u| = (1/4) ln [e^(2x)/(e^(2x) + 4)] + C

 

[fasteddie65]

Re: Integrate 2/(e^(2x) + 4) dx using substitution TYPO ALERT!

TYPO ALERT! In the last line, the ? symbol should be a ? symbol instead.

 

[Monkeyseat]

Thanks for replying. A couple of things:

1) I have no idea what "Partial Fractional Decomposition" is.

2) My working is probably wrong, but the answer in the book is:

(x/2) - (1/4)ln(e^(2x) + 4) + c

Please could someone clear this up... I am really confused.

Thanks.

 

[soroban]

Hello, Monkeyseat!

Your work is correct . . .

\(\displaystyle \int\frac{2}{e^{2x}+4}\,dx\)

\(\displaystyle \text{Let }\,u \:=\:e^{2x}+4\)

\(\displaystyle \text{Substitute: }\:\int\frac{du}{u(u-4)}\)

You say you don't know Partial Fractions?
Well, we can still integrate it . . . it takes some extra work, though.


\(\displaystyle \text{The denominator is: }\;u^2 - 4u\)

\(\displaystyle \text{Add and subtract 4: }\;u^2 - 4u + 4 - 4 \;=\;(u-2)^2 - 4\)

\(\displaystyle \text{The integral becomes: }\;\int\frac{du}{(u-2)^2 - 4}\)

\(\displaystyle \text{Let: }\,u \:=\:2\sec\theta \quad\Rightarrow\quad du \:=\:2\sec\theta\tan\theta\,d\theta\)

\(\displaystyle \text{Substitute: }\;\int\frac{2\sec\theta\tan\theta\,d\theta}{4\tan^2\!\theta} \quad\Rightarrow\quad \tfrac{1}{2}\!\!\int\csc\theta\,d\theta \;=\;\tfrac{1}{2}\ln(\csc\theta - \cot\theta) + C\) .[1]


\(\displaystyle \text{Back-substitute . . .}\)
. . \(\displaystyle \text{We have: }\:u-2 \:=\:2\sec\theta \quad\Rightarrow\quad \sec\theta \:=\:\frac{u-2}{2} \:=\:\frac{hyp}{adj}\)

We have this right triangle:

                  *
        (u-2)  *  |
            *     |
         * @      |
      * - - - - - *
            2

\(\displaystyle \text{Since }adj = 2,\;hyp = u-2\text{, we have: }\pp = \sqrt{u^2-4u}\)

\(\displaystyle \text{Then: }\:\csc\theta \:=\:\frac{hyp}{opp}\:=\:\frac{u-2}{\sqrt{u^2-4u}},\;\;\cot\theta \:=\:\frac{adj}{opp} \:=\:\frac{2}{\sqrt{u^2-4}}\)


Substitute into [1]:

. . \(\displaystyle \frac{1}{2}\ln\left(\frac{u-2}{\sqrt{u^2-4u}} - \frac{2}{\sqrt{u^2-4u}}\right) + C \;=\;\frac{1}{2}\ln\left(\frac{u-4}{\sqrt{u(u-4)}}\right) + C\)


\(\displaystyle \text{Back-substitute: }\:u \:=\:e^{2x} + 4 \quad\Rightarrow\quad u - 4 \:=\:e^{2x}\)

\(\displaystyle \text{We have: }\;\frac{1}{2}\ln\left[\frac{e^{2x}}{\sqrt{(e^{2x}+4)e^{2x}}}\right] + C \;=\;\frac{1}{2}\ln\left[\frac{e^{2x}}{e^x\sqrt{e^{2x}+4}}\right] + C \;=\;\frac{1}{2}\ln\left[\frac{e^x}{\sqrt{e^{2x}+4}}\right] + C\)

. . \(\displaystyle = \;\frac{1}{2}\bigg[\ln(e^x) - \ln\left(\sqrt{e^{2x}+4}\right)\bigg] + C \;=\;\frac{1}{2}\bigg[x - \frac{1}{2}\ln\left(e^{2x} + 4\right)}\bigg] + C\)

. . \(\displaystyle = \;\frac{x}{2} - \frac{1}{4}\ln(e^{2x} + 4) + C\quad\hdots\) . ta-DAA!

 

[Monkeyseat]

Sorry for the late response.

Thanks both of you very much. I read up on "Partial Fractional Decomposition" so I understand the first method now also.

Thanks again.