Evaluate $\int \frac{e^{-2x}}{e^{-2x}-3}dx$ $$\displaystyle\int \dfrac{e^{-2x}}{e^{-2x}-3}dx$$
I'm not sure how to integrate this. What's the first step? I thought it was the common result where the numerator is the derivative of the denominator, but that doesn't seem to be the case here.
 A: Remember that $(e^{-2x} - 3)' = -2e^{-2x}$. So, just multiply and divide by $-2$:
$$\int \frac{-2}{-2} \cdot \dfrac{e^{-2x}}{e^{-2x}-3} dx =$$
$$-\frac{1}{2} \int \dfrac{-2e^{-2x}}{e^{-2x}-3} dx.$$
Now the numerator is the derivative of the denominator. So, the solution is:
$$-\frac{\ln{|e^{-2x} - 3|}}{2} + C,$$
where $C$ is a constant.
A: $$let\\u=e^{-2x}-3\\du=-2e^{-2x}dx\\\frac{-1}{2}du=e^{-2x}dx\\\int \frac{e^{-2x}}{e^{-2x}-3}dx=\\\frac{-1}{2}\int \frac{1}{u}du=\\\frac{-1}{2}ln(u)+c
$$
A: Instead of doing that $u=\dots\frac{du}{dx}=\dots dx=\dots$ thingy, we can perform a substitution more directly just by multiplying by one. Keep an eye on what happens to the derivative of the substitution.
$$\begin{array}{lll}
\int\frac{e^{-2x}}{e^{-2x}+3}dx&=&\int\frac{e^{-2x}}{e^{-2x}+3}dx\cdot 1\\
&=&\int\frac{e^{-2x}}{e^{-2x}+3}\color{green}{dx}\cdot \frac{\frac{d(e^{-2x+3})}{\color{green}{dx}}}{\color{blue}{\frac{d(e^{-2x+3})}{dx}}}\\
&=&\int\frac{e^{-2x}}{e^{-2x}+3}\color{green}{dx}\cdot \frac{\frac{d(e^{-2x+3})}{\color{green}{dx}}}{\color{blue}{-2e^{-2x}}}\\
&=&\int\frac{e^{-2x}}{\color{blue}{-2e^{-2x}}(e^{-2x}+3)}\color{green}{dx}\cdot \frac{d(e^{-2x+3})}{\color{green}{dx}}\\
&=&\int\frac{1}{-2(e^{-2x}+3)} d(e^{-2x+3})\\
&=&\int\frac{1}{-2u} du\\
&=&\dots
\end{array}$$
Did you notice that when we integrate an expression, and perform a substitution, we divide the expression by the derivative of the substitution?
It may be instructive to do a few substitutions this way, as well as the $u=\dots\frac{du}{dx}=\dots dx=\dots$ thingy way so you can compare them. The reason why I prefer the way that I demonstrated here is because I didn't have to format the numerator (e.g. $e^{-2x}=\frac{1}{-2}\cdot(-2e^{-2x})$) before substituting. I just performed a cancellation right off the bat.
Note that throughout, I've referred to the two techniques as "my way" or the "thingy way" because I wanted to avoid calling them different methods. This is because they are actually the same method.
A: Hint:
Set $u=e^{-2x}$ then $dx=-\frac{du}{2u}$ and $$\int...=\int\frac{-u}{2u(u-3)}=...$$
A: Directly: since
$$\int\frac{f'(x))}{f(x)}dx=\log f(x) + C$$
we get
$$\frac{e^{-2x}}{e^{-2x}-3}=-\frac12\frac{-2e^{-2x}}{e^{-2x}-3}=-\frac12\frac{\left(e^{-2x}-3\right)'}{e^{-2x}-3}\implies$$
$$\int\frac{e^{-2x}}{e^{-2x}-3}dx=-\frac12\log\left(3^{-2x}-3\right)+C$$
