# Thread: Integration by substitution involving e^2x + 4

1. ## Integration by substitution involving e^2x + 4

Hi everyone,

I am trying to integrate the following expression:

$\displaystyle \int$$\displaystyle \frac{2}{u}$$\displaystyle \mathrm{d}x$

(where "u" is it should read "$\displaystyle e^{2x}$$\displaystyle +4" but I couldn't get the LaTex maths thing to work, sorry) using the given substitution \displaystyle u = \displaystyle e^{2x}$$\displaystyle +4$

Please can anyone offer a way to do this? I get to the stage where I have:

$\displaystyle \int$$\displaystyle \frac{1}{a}$$\displaystyle du$

(where "a" is it should read "u(u-4)")

but I can't get any further. Any help would be much appreciated.

2. ## Re: Integration by substitution involving e^2x + 4

If you have to integrate:
$\displaystyle \int \frac{2dx}{e^{2x}+4}$

If you say, let $\displaystyle u=e^{2x}+4 \Rightarrow du=2\cdot e^{2x}dx \Leftrightarrow \frac{du}{u-4}=2dx$, you'll get:
$\displaystyle \int \frac{2dx}{e^{2x}+4}=\int \frac{du}{u\cdot (u-4)}$

Try to split up in partial fractions:
$\displaystyle \frac{1}{u\cdot(u-4)}=\frac{A}{u}+\frac{B}{u-4}$

Determine A and B.

3. ## Re: Integration by substitution involving e^2x + 4

Originally Posted by Lex
$\displaystyle \int$$\displaystyle \frac{2}{u}$$\displaystyle \mathrm{d}x$

(where "u" is it should read "$\displaystyle e^{2x}$$\displaystyle +4$" but I couldn't get the LaTex maths thing to work, sorry)
Just notice that $\displaystyle \frac{1}{e^{2x}+4}=\frac{e^{-2x}}{1+4e^{-2x}}$.
Now it is easy.

4. ## Re: Integration by substitution involving e^2x + 4

Siron, I got $\displaystyle \frac{1}{u\cdot(u-4)}=\frac{1}{4u-16}-\frac{1}{4u}$

If this is correct, I can't integrate the $\displaystyle \frac{1}{4u-16}$, can I?

5. ## Re: Integration by substitution involving e^2x + 4

Sure you can: $\displaystyle \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C$

It may be easier to write $\displaystyle \dfrac{1}{4u+16} = \dfrac{1}{4} \cdot \dfrac{1}{u+4}$

6. ## Re: Integration by substitution involving e^2x + 4

Originally Posted by Lex
I got $\displaystyle \frac{1}{u\cdot(u-4)}=\frac{1}{4u-16}-\frac{1}{4u}$

If this is correct, I can't integrate the $\displaystyle \frac{1}{4u-16}$, can I?
Why not? since the denominator is linear, the substitution v= 4u- 16 is easy.
(or write the fraction as $\displaystyle \frac{1}{4}\frac{1}{u- 4}$ and use the substitution v= u- 4.)

7. ## Re: Integration by substitution involving e^2x + 4

Originally Posted by Plato
Just notice that $\displaystyle \frac{1}{e^{2x}+4}=\frac{e^{-2x}}{1+4e^{-2x}}$.
Now it is easy.
I used the substitution $\displaystyle u = 1 + 4e^{-2x}$ and got:

$\displaystyle -\frac{1}{8}ln(e^{2x}+4) + c$

My book says the answer is $\displaystyle \frac{1}{2}x - \frac{1}{4}ln(e^{2x}+4) + c$

I'm a bit confused!

8. ## Re: Integration by substitution involving e^2x + 4

Originally Posted by Lex
I used the substitution $\displaystyle u = 1 + 4e^{-2x}$ and got: $\displaystyle -\frac{1}{8}ln(e^{2x}+4) + c$
My book says the answer is $\displaystyle \frac{1}{2}x - \frac{1}{4}ln(e^{2x}+4) + c$ I'm a bit confused!
I second that.

If the problem is $\displaystyle \int {\frac{{dx}}{{e^{2x} + 4}}}$ then the answer is $\displaystyle \frac{1}{4}x - \frac{1}{8}\ln \left( {e^{2x} + 4} \right)+c$.

9. ## Re: Integration by substitution involving e^2x + 4

OK...

I get the right answer (or at least the same answer as the book) when I use $\displaystyle \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C$

I also get the right answer when I use the substitution v = 4u - 16.

So thanks to e^(i*pi) and HallsofIvy.

As for using the substitution $\displaystyle u = 1 + 4e^{-2x}$ and Plato's answer, I have no idea what's going on!!

10. ## Re: Integration by substitution involving e^2x + 4

Originally Posted by Lex
OK...

I get the right answer (or at least the same answer as the book) when I use $\displaystyle \int \dfrac{1}{ax+b}\ dx = \dfrac{1}{a} \ln |ax+b| + C$

I also get the right answer when I use the substitution v = 4u - 16.

So thanks to e^(i*pi) and HallsofIvy.

As for using the substitution $\displaystyle u = 1 + 4e^{-2x}$ and Plato's answer, I have no idea what's going on!!
Read Plato's post again. Look at the numerator of the integrand. So multiply his answer by 2 (I think Plato misread the question - an easy mistake to make given the confusing way the questin was posted).

11. ## Re: Integration by substitution involving e^2x + 4

OK, yes, I see that the 2 was missing.

Way to patronise.

12. ## Re: Integration by substitution involving e^2x + 4

Originally Posted by Lex
OK, yes, I see that the 2 was missing.

Way to patronise.
Excuse me??

Multiply the numerator and denominator of your integrand by e^(-2x) to get what Plato has. Then make the substitution u = 1 + 4e^(-2x) and the answer is immediate. ln|1 + 4 e^(-2x)| and ln|e^(2x) + 4)|-2x are equivalent by the way ...... (Use basic log rules to prove this).

13. ## Re: Integration by substitution involving e^2x + 4

An alternative:

\displaystyle \displaystyle \begin{align*} \int{\frac{2}{e^{2x} + 4}\,dx} &= \int{\frac{2e^x}{e^{3x} + 4e^x}\,dx} \\ &= \int{\frac{2}{u^3 + 4u}\,du}\textrm{ after making the substitution }u = e^x \implies du = e^x\,dx \\ &= \int{\frac{2}{u(u^2 + 4)}\,du} \end{align*}

Now applying Partial Fractions

\displaystyle \displaystyle \begin{align*} \frac{A}{u} + \frac{Bu + C}{u^2 + 4} &= \frac{2}{u(u^2 + 4)} \\ \frac{A(u^2 + 4) + (Bu + C)u}{u(u^2 + 4)} &= \frac{2}{u(u^2 + 4)} \\ A(u^2 + 4) + (Bu + C)u &= 2 \\ Au^2 + 4A + Bu^2 + Cu &= 2 \\ (A + B)u^2 + Cu + 4A &= 0u^2 + 0u + 2 \end{align*}

So $\displaystyle \displaystyle 4A = 2 \implies A = \frac{1}{2}, C = 0, A + B = 0 \implies \frac{1}{2} + B = 0 \implies B = -\frac{1}{2}$.

Therefore $\displaystyle \displaystyle \frac{\frac{1}{2}}{u} + \frac{-\frac{1}{2}u}{u^2 + 4} = \frac{2}{u(u^2 + 4)}$

So completing the integral...

\displaystyle \displaystyle \begin{align*}\int{\frac{2}{u(u^2 + 4)}\,du} &= \frac{1}{2}\int{\frac{1}{u}\,du} - \frac{1}{2}\int{\frac{1}{u^2 + 4}\,du} \\ &= \frac{1}{2}\ln{|u|} - \frac{1}{2}\arctan{\left(\frac{u}{2}\right)} +C \\ &= \frac{1}{2}\ln{\left|e^x\right|} - \frac{1}{2}\arctan{\left(\frac{e^x}{2}\right)} +C \\ &= \frac{x}{2} - \frac{1}{2}\arctan{\left(\frac{e^x}{2}\right)} + C\end{align*}

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