# 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.

• let the denominator be t, differentiate both sides, u may get something – Shobhit Sep 2 '14 at 18:27
• $$\frac{d[e^{-2x}-3]}{dx}=?$$ – lab bhattacharjee Sep 2 '14 at 18:27
• Apart from a multiplicative constant, it is the case here. – André Nicolas Sep 2 '14 at 18:29
• "where the numerator is the derivative of the denominator" isn't exactly correct. Quite often the derivative is a factor of the substitution. Consider $$\int \frac{5x}{(x^2+1)(x^2+2)}dx$$ the obvious substitution is $u=x^2$, but $2x$ is not the numerator, and $x^2$ is not the denominator.Try to express $5x$ as $\frac{5}{2}\cdot 2x$ – John Joy Sep 2 '14 at 23:14

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.

$$let\\u=e^{-2x}-3\\du=-2e^{-2x}dx\\\frac{-1}{2}du=e^{-2x}dx\\\int \frac{e^{-2x}}{e^{-2x}-3}dx=\\\frac{-1}{2}\int \frac{1}{u}du=\\\frac{-1}{2}ln(u)+c$$

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.

Hint:

Set $u=e^{-2x}$ then $dx=-\frac{du}{2u}$ and $$\int...=\int\frac{-u}{2u(u-3)}=...$$

• It should be a $-3$ in the denominator. – Tomás Sep 2 '14 at 18:50
• Tks, I correct it. But anyway, there is other answers :-) – idm Sep 2 '14 at 18:52

Directly: since

$$\int\frac{f'(x))}{f(x)}dx=\log f(x) + C$$

we get

$$\frac{e^{-2x}}{e^{-2x}-3}=-\frac12\frac{-2e^{-2x}}{e^{-2x}-3}=-\frac12\frac{\left(e^{-2x}-3\right)'}{e^{-2x}-3}\implies$$

$$\int\frac{e^{-2x}}{e^{-2x}-3}dx=-\frac12\log\left(3^{-2x}-3\right)+C$$