I'm trying to figure out how to integrate this function. I tried several tricks from my toolkit, but I can't seem to figure it out.

$$\int\ \frac {e^{2x}-6e^x}{e^x+2}\ dx$$

So let's say that I factor out some terms and split the equation to:

$\displaystyle\int\ \frac {e^{2x}}{e^x+2}\ dx\;$ and $\;\displaystyle -6\int\ \frac {e^x}{e^x+2}\ dx$

Now I cannot see how the derivative of $e^{2x}$ is $e^x+2$ or vice versa. When looking at the other side, it seems more reasonable to set $u$ to $e^x$. So let's do that:

$\displaystyle -6\int\ \frac {e^x}{e^x+2}\ dx$ = $\displaystyle -6\int\ \frac {u}{u+2}\ dx$

Okay, so as you can see, I didn't get anywhere here. I'm really new to $u$-substitution and partial integration, I probably missed some crucial step.


$$\int\ \frac {e^{2x}-6e^x}{e^x+2}\ dx$$

Let $\displaystyle e^x + 2 = u$.$^{\color{blue}{(1)}}$

So $\displaystyle e^x = u-2$, and so $\displaystyle e^{2x} = (u-2)^2$.

$\color{blue}{(1)}$ Now $\displaystyle e^x \ dx = du \iff \ dx = \frac{du}{e^x}$, but recall from above, that $\displaystyle e^x = u-2$.

So in fact, $\displaystyle \color{blue}{dx = \frac{du}{u-2}}$.

That gives us the integral

$$\begin{align} \int \frac{(u-2)^2 - 6(u-2)}{u\cdot \color{blue}{(u-2)}}\ \color{blue}{du} & = \int \frac{(u-2) - 6}{u}\,du \\ & = \int \frac{u-8}{u}\,du \\ &= \int \left(1- \frac 8u\right)\ du = \end{align}$$

Can you take it from here?

  • $\begingroup$ Thanks! I get it now :). $\endgroup$ – user472288 May 11 '18 at 23:43
  • $\begingroup$ Your welcome! ${}$ $\endgroup$ – Namaste May 11 '18 at 23:46

The trick is to see that $e^{2x}$ can be easily changed and factored. We start with the integral. $$\int \frac {e^{2x}-6e^x}{e^x+2}~ {\rm d}x$$ We can use the fact that $a^{mn} = (a^m)^n$ to change this to $$\int \frac {(e^x)^2-6e^x}{e^x+2}~ {\rm d}x = \int \frac {e^x(e^x - 6)}{e^x+2}~ {\rm d}x$$ We can then use the substitution $u = e^x +2$ and ${\rm d}u = e^x{\rm d}x$ and solve from there. $$\int \frac {u - 8}{u}~ {\rm d}u = \int \left(1 - \frac{8}{u}\right){\rm d}u = u - 8\ln|u| + C$$ We then substitute the equivalent of $u$. $$e^x + 2 - 8\ln|e^x + 2| + C$$ We can also now say that the $2$ can become part of the constant $C$ and remove the absolute value, since $e^x + 2 > 0$ for $x\in\mathbb{R}$ giving us a final answer of $$e^x - 8\ln\left(e^x + 2\right) + C$$


$$\int{\frac{e^{2x}-6e^x}{e^x+2}dx}=\int{\frac{e^{2x}}{e^x+2}dx}-6\int{\frac{e^x}{e^x+2}dx}$$ Substitution: $u=e^x+2$, then $\frac{du}{dx}=e^x=u-2\to dx=\frac{1}{u-2}du$

Sub into first half of integral for:


And second half of integral will get:


Thus overall you get: $$u-2\ln|u|-6\ln|u| +C=u-8\ln|u|+C$$


Hint: $$\frac{e^{2x}-6e^x}{e^x+2}= \frac{e^{2x}}{e^x+2}-\frac{6e^x}{e^x+2}$$

Further hint: Let $u=e^x+2$ then $dx=\frac{du}{e^x}$

So when you substitute this back in the first one, it becomes $\int\frac{u-2}{u}du$. Double-check this carefully.

  • $\begingroup$ Thanks Pure, that does make sense but what about the left side? $u$ at $e^x+2$ would result in a derivative that is not $e^{2x}$ $\endgroup$ – user472288 May 11 '18 at 23:13
  • $\begingroup$ It does not result in that derivative, however it does not need to, you still get a cancellation $\endgroup$ – pureundergrad May 11 '18 at 23:16
  • $\begingroup$ Your problem is that you didn't replace $dx$ with $du/e^x$, once you do that you are nearly done $\endgroup$ – pureundergrad May 11 '18 at 23:19
  • 1
    $\begingroup$ pureundergrad $$u = e^x+2 \iff e^x = u-2 \iff e^{2x} = (u-2)^2$$ Also, dx is expressed in terms of both x and du. That might be confusing. OP: But you do need to use $$du = e^x dx \iff dx = \frac {du}{e^x} = \frac{du}{u-2}$$ So when substituting, where you see $dx$ in the integral, replace it with $\frac {du}{u-2}$ $\endgroup$ – Namaste May 11 '18 at 23:25

It looks to me like you are almost done. You have reduced the problem to integrating $-6\int \frac{e^x}{e^x+ 2}dx$. Now let $u= e^x+ 2$. Then $du= e^xdx$ so that integral becomes $-6\int \frac{1}{u}du$.


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.