1
$\begingroup$

$$\displaystyle\int \dfrac{e^{-2x}}{e^{-2x}-3}dx$$

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.

$\endgroup$
  • $\begingroup$ let the denominator be t, differentiate both sides, u may get something $\endgroup$ – Shobhit Sep 2 '14 at 18:27
  • 1
    $\begingroup$ $$\frac{d[e^{-2x}-3]}{dx}=?$$ $\endgroup$ – lab bhattacharjee Sep 2 '14 at 18:27
  • 3
    $\begingroup$ Apart from a multiplicative constant, it is the case here. $\endgroup$ – André Nicolas Sep 2 '14 at 18:29
  • $\begingroup$ "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 $\frac{5}{2}\cdot 2x$ $\endgroup$ – John Joy Sep 2 '14 at 23:14
5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Hint:

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

$\endgroup$
  • $\begingroup$ It should be a $-3$ in the denominator. $\endgroup$ – Tomás Sep 2 '14 at 18:50
  • $\begingroup$ Tks, I correct it. But anyway, there is other answers :-) $\endgroup$ – idm Sep 2 '14 at 18:52
0
$\begingroup$

Directly: since

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

we get

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.