Here's the integral:

$$\int\frac{x e^{2x}}{(1+2x)^2}dx$$

Every method I try to use either hits a dead end or makes the problem more complicated. The only way I've managed to actually complete the integral is using integration by parts and distributing everything all the way out which but that ends up being a hugely complex mess of different terms that I have no way of verifying.


The answer in the back of the book and the answer given by Mathematica is:


However I can't seem to get there. I feel like there must be a way to arrange the completed integral algebraically so that all the complicated terms cancel and you end up with the nice clean answer the book gives, but after an hour of trial and error I haven't found it yet.

Edit: Apologies I did make a typo in this post, it has been corrected but I didn't make that error when computing the problem.

To confirm (for you and myself):

enter image description here enter image description here

  • 1
    $\begingroup$ There is a typo in the book's problem. If the negative sign is changed to a positive sign the answer quoted by the book works. $\endgroup$ – Leucippus Mar 14 '15 at 22:49
  • $\begingroup$ I did mistype that when entering the problem here (I just corrected it), but Not when I was computing the problem on paper. $\endgroup$ – user3776749 Mar 14 '15 at 22:51


The answer in the back of your book is wrong, unless you mistyped something here. Here is a graph of the derivatives of both functions. You can see that they are not equal. They don't even have the same singularities.

enter image description here

If you change the question to

$$\int\frac{x e^{2x}}{(2x+1)^2}\,dx$$

then the answer in the back of the book is correct. Do you need help with that problem?


Apparently you do want help with the corrected problem. When you are stuck on doing an integration and you know the correct answer, a clue is to write out taking the derivative of the answer and getting the integrand of your question. You can then do that all backward.

If that makes the final answer seem too "magic," as it does here, replace the reverse-of-the-product-rule with an integration by parts.

Here is the method the book probably wants you to use: Do integration by parts, with $e^{2x}$ as the $u$ and the rest as $dv$.

Do you need more help?

  • $\begingroup$ @Amzoti: TI-Nspire Teacher Software, running as a TI-Nspire CX. I sometimes use the free software Geogebra, but I have used the TI software more and am more comfortable with it. When I have more time I crop out the top border of the graphic. $\endgroup$ – Rory Daulton Mar 14 '15 at 23:06
  • $\begingroup$ On HELP FOR THE REVISED QUESTION. Give me a brief moment to work through that. $\endgroup$ – user3776749 Mar 14 '15 at 23:12
  • 1
    $\begingroup$ I got there, eventually. After some frankly labyrinthine algebra. Thanks for that tip, it was extremely helpful and I'll be certain to use it in the future. Though I must say I dislike problems where the only option is to venture blindly forth hoping it all works out in the end :) $\endgroup$ – user3776749 Mar 14 '15 at 23:42

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.