6
$\begingroup$

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

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?

$\endgroup$
1
  • 3
    $\begingroup$ 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 $\operatorname d v=\frac{\operatorname d x}{(1+2x)^2} \implies v = \frac{-1}{2(1+2x)}+c$ $\endgroup$
    – Graham Kemp
    Jan 29 '16 at 22:41
7
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Don't really know who downvoted and why o.O I just gave you an UP for the simplicity of the answer ^^ $\endgroup$
    – Turing
    Jan 29 '16 at 22:43
5
$\begingroup$

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

https://en.wikipedia.org/wiki/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!

$\endgroup$
3
  • $\begingroup$ @zz20s Whoops.. I thought it was just a possibility, not a requirement! $\endgroup$
    – Turing
    Jan 29 '16 at 22:43
  • 1
    $\begingroup$ 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! $\endgroup$
    – zz20s
    Jan 29 '16 at 22:44
  • $\begingroup$ @zz20s Wow thank you! I should have read more carefully :D $\endgroup$
    – Turing
    Jan 29 '16 at 22:45
2
$\begingroup$

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

$\endgroup$

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.