Integration by Substitution of Fraction involving e

Question

Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$

The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$

I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.

$\int\frac{2}{e^{2x}+4}$

let $u = e^{2x} +4$

$\frac{dy}{dx}=2e^{2x}$

$dx=\frac{1}{2}e^{-2x}du$

$u = e^{2x} +4$

$e^{2x} = u-4$

$e^{-2x} = \frac{1}{u-4}$

$dx=\frac{1}{2}(\frac{1}{u-4})du$

Hence the integral is:

$$ \begin{aligned} &\int\frac{2}{e^{2x}+4} \\ =& 2\int\frac{1}{u}\frac{1}{2}\left(\frac{1}{u-4}\right)du\\ =&\int\frac{1}{u}\left(\frac{1}{u-4}\right)du\\ =&\int\frac{1}{u^2-4u}du\\ =&\int u^{-2}-\frac{1}{4}u^{-1}du\\ =&\frac{u^{-1}}{-1}-\frac{1}{4}\ln(u)+c\\ =&-\frac{1}{e^{2x}+4}-\frac{1}{4}\ln(e^{2x}+4)+c\\ =&-e^{-2x}-\frac{1}{4}-\frac{1}{4}\ln(e^{2x}+4)+c\\ \ne& \frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c\\ \end{aligned} $$

???? help

Answer 1

You have made a basic, but common algebraic error:

$$\int\frac{1}{u^2-4u}du \neq\int u^{-2}-\frac{1}{4}u^{-1}du$$

You cannot split denominator like that. To continue your method, use partial fractions.

Answer 2

You wrote $$\int\frac{1}{u^2-4u}du=\int u^{-2}-\frac{1}{4}u^{-1}du$$ Which is incorrect

Try writing it as $$\int\frac{1}{u^2-4u}du=\frac{1}{4}\int \frac{1}{u-4}-\frac{1}{u}du$$

Or

$$\int\frac{1}{u^2-4u}du=\int\frac{1}{(u-2)^2-4}du$$ And proceed through substitution

Answer 3

You have made a mistake:

Note that $$\int\frac{1}{u^2-4u}du \ne \int u^{-2}-\frac{1}{4}u^{-1}du$$

By the way $$\int\frac{2dx}{e^{2x}+4} = \int\frac{2e^{2x}dx}{e^{2x}(e^{2x}+4)} = $$

$$\int\frac{du}{u(u+4)}$$

Where $$u= e^{2x}$$

is a short way to go.