Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


With this particular problem. my approach is to to rewrite the integral as $$\int xe^{2x}\frac{1}{(1+2x)^2}\,dx$$ and then pick a $u$ and a $dv$ and take it from there. The only issue I'm running into is that $xe^{2x}$ appears to me as two functions instead of one. What is a suggestion for this?

share|cite|improve this question
Hint: There's no real issue with that choice; it's ideal. $\;$ We have $u=x {\sf e}^{2x} \implies \operatorname d u = (1+2x){\sf e}^{2x}\operatorname d x$ via the product rule. $\;$ Which is rather convenient when coupled with $\operatorname d v=\frac{\operatorname d x}{(1+2x)^2} \implies v = \frac{-1}{2(1+2x)}+c$ – Graham Kemp Jan 29 at 22:41
up vote 7 down vote accepted

I would suggest letting $u=xe^{2x}$, since it must be integrated by parts (whereas the derivative is the product rule), and $dv=\frac{1}{(1+2x)^2}dx$.

You can allow a composite of two functions be equal to $u$. That is perfectly valid and often required.

share|cite|improve this answer
Don't really know who downvoted and why o.O I just gave you an UP for the simplicity of the answer ^^ – Time Master Jan 29 at 22:43

Make the substitution $1 + 2x = t$ so that $\text{d}x = \frac{\text{d}t}{2}$ so you get:

$$I = \int \frac{(t-1)}{2}\cdot \frac{e^{2\left(\frac{(t-1)}{2}\right)}}{t^2}\frac{\text{d}t}{2}$$

and arranging the terms you easily get:

$$\frac{e^{-1}}{4}\int \frac{e^t}{t} - \frac{e^{t}}{t^2}\ \text{d}t$$

Now the first integral is a Special Function called the Exponential Integral function:

$$\int\frac{e^t}{t}\ \text{d}t = \text{Ei}(t)$$

and the second one can be performed by parts, giving the quite same result:

$$\int \frac{e^t}{t^2}\ \text{d}t = -\frac{e^t}{t} + \text{Ei}(t)$$

Putting together and you see the two Special Functions are cancelled by the minus sign, obtaining in the end the result of the integration in $\text{d}t$:

$$\frac{e^{-1}}{4}\frac{e^{t}}{t} \equiv \frac{e^{t-1}}{4t}$$

coming back to $x$:

$$I = \frac{e^{2x}}{4\cdot(1 + 2x)}$$

More about Exponential Integral

Final Remark

Don't forget about the various $C$ constants you can obtain from each integration, and you can set up them as zero!

share|cite|improve this answer
@zz20s Whoops.. I thought it was just a possibility, not a requirement! – Time Master Jan 29 at 22:43
No problem at all. +1 for an ingenious solution. It's not a requirement, but since the OP leaved towards parts, I just wanted to make sure you knew! – zz20s Jan 29 at 22:44
@zz20s Wow thank you! I should have read more carefully :D – Time Master Jan 29 at 22:45

To use the integration by parts method, we do the following: $$\int \frac{1}{(1+2x)^2}xe^{2x}dx=\int \left (-\frac{1}{2(1+2x)}\right )'xe^{2x}dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\int \frac{1}{2(1+2x)}(e^{2x}+2xe^{2x})dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\int \frac{1}{2(1+2x)}(e^{2x}(1+2x))dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\int \frac{e^{2x}}{2}dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\frac{e^{2x}}{4}+C \\ =e^{2x}\frac{1+2x-2x}{4(1+2x)}+C\\ =\frac{e^{2x}}{4(1+2x)}+C$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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