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

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?

Martin Sleziak
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wa7d
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    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 '16 at 22:41

3 Answers3

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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.

zz20s
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    Don't really know who downvoted and why o.O I just gave you an UP for the simplicity of the answer ^^ –  Jan 29 '16 at 22:43
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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!

  • @zz20s Whoops.. I thought it was just a possibility, not a requirement! –  Jan 29 '16 at 22:43
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    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 '16 at 22:44
  • @zz20s Wow thank you! I should have read more carefully :D –  Jan 29 '16 at 22:45
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To use the integration by parts method, we do the following: $$\int \frac{1}{(1+2x)^2}xe^{2x}dx=\int \left (-\frac{1}{2(1+2x)}\right )'xe^{2x}dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\int \frac{1}{2(1+2x)}(e^{2x}+2xe^{2x})dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\int \frac{1}{2(1+2x)}(e^{2x}(1+2x))dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\int \frac{e^{2x}}{2}dx \\ =-\frac{1}{2(1+2x)}xe^{2x}+\frac{e^{2x}}{4}+C \\ =e^{2x}\frac{1+2x-2x}{4(1+2x)}+C\\ =\frac{e^{2x}}{4(1+2x)}+C$$

Mary Star
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