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$\int_{} \frac{xe^{2x}}{(1+2x)^2}dx$

I am having trouble picking the correct $u/dv$ before integrating by parts. I felt like L.I.A.T.E. did not really help me here...

This is what I tried, but it ended up with integration spiraling into an endless evaluation of an integral...

$$ \begin{align} u &= (1 + 2x)^2 & dv &= xe^{2x}dx \\ du &= 2(1+2x)dx & v &= \frac{1}{2}x^2 \frac{1}{2}e^{2x} \\ &= 2 + 4xdx & &= \frac{1}{4}x^2e^{2x} \end{align} $$

Am I at least correct in choosing the right $u/du$ values? That is all I really want to know, if I am allowed to choose the $u/du$ like how I did

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  • $\begingroup$ Check $du$. You've made an error when applying the chain rule. Recall: $$[(f(x))^n]'=nf'(x)(f(x))^{n-1}$$ $\endgroup$ – Rhys Hughes Feb 18 at 21:56
  • $\begingroup$ oops, you're right. Should be $du = 1(1+2x)dx$, thanks! $\endgroup$ – Evan Kim Feb 18 at 22:02
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    $\begingroup$ @EvanKim Actually, it should be $du=4(1+2x)dx$. $\endgroup$ – Haris Gusic Feb 18 at 22:04
  • $\begingroup$ oh, yes you're right. I am not thinking right anymore lol, time to take a break $\endgroup$ – Evan Kim Feb 18 at 22:05
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Hint:

Let $u=xe^{2x}$, $\text dv=\dfrac{\text dx}{(1+2x)^2}$.

Result: $$\dfrac{\mathrm{e}^{2x}}{8x+4} + C$$

EDIT: More steps.

\begin{eqnarray*} \int \frac{xe^{2x}}{(1+2x)^2} \ \text dx &=& \left|\begin{array}{2} u=xe^{2x} & \text dv=\dfrac{\text d x}{(1+2x)^2} \\ \text du = (1+2x)e^{2x}\ \text dx & v=-\dfrac{1}{2(1+2x)} \end{array}\right| = \\ &=& -\frac{xe^{2x}}{2(1+2x)} + \frac{1}{2} \int e^{2x}\ \text dx = \\ &=&-\frac{xe^{2x}}{2(1+2x)} + \frac{1}{4}e^{2x} + C = \\ &=& e^{2x}\left( -\frac{x}{2(1+2x)}+\frac{1}{4} \right) + C = \boxed{\frac{e^{2x}}{8x+4} + C} \end{eqnarray*}

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  • $\begingroup$ Hey, I know you answered this almost a week ago...I have tried for HOURS to try to work this out given your hint and have not been successful yet :( $\endgroup$ – Evan Kim Feb 25 at 17:03
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    $\begingroup$ @EvanKim There you go. I added more steps. $\endgroup$ – Haris Gusic Feb 25 at 17:23
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    $\begingroup$ @EvanKim $$\cdots=\frac{-2x+1+2x}{4(1+2x)}$$ $\endgroup$ – Haris Gusic Feb 25 at 19:10
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    $\begingroup$ Thank you, your whole process seemed like a good way of getting through all of the spots that were tricky to me (and also filled in a lot of my blank spots to make more sense) $\endgroup$ – Evan Kim Feb 25 at 19:25
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    $\begingroup$ @EvanKim You are welcome! $\endgroup$ – Haris Gusic Feb 25 at 19:29
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$u=1+2x$:

$$\require{cancel} \int\frac{xe^{2x}}{(1+2x)^2}\,dx= \frac{1}{4}\int\frac{(1+2x-1)e^{2x+1-1}}{(1+2x)^2}\frac{d}{dx}(1+2x)\,dx=\\ \frac{1}{4}\int\frac{(u-1)e^{u-1}}{u^2}\,du= \frac{1}{4e}\int\left(\frac{ue^{u}}{u^2}-\frac{e^{u}}{u^2}\right)\,du=\\ \frac{1}{4e}\int\left(\frac{e^{u}}{u}-\frac{e^{u}}{u^2}\right)\,du= \frac{1}{4e}\left(\int e^{u}\frac{1}{u}\,du-\int e^{u}\frac{1}{u^2}\,du\right)=\\ \frac{1}{4e}\left(\int e^{u}(\ln{u})'\,du+\int e^{u}\left(\frac{1}{u}\right)'\,du\right)=\\ \frac{1}{4e}\left(e^{u}\ln{u}-\int e^{u}\ln{u}\,du+\frac{e^u}{u}-\int e^{u}\frac{1}{u}\,du\right)=\\ \frac{1}{4e}\left(e^{u}\ln{u}-\int e^{u}\ln{u}\,du+\frac{e^u}{u}-\int e^{u}(\ln{u})'\,du\right)=\\ \frac{1}{4e}\left(\cancel{e^{u}\ln{u}}-\cancel{\int e^{u}\ln{u}\,du}+\frac{e^u}{u}-\cancel{e^{u}\ln{u}}+\cancel{\int e^{u}\ln{u}\,du}\right)=\\ \frac{1}{4e}\frac{e^u}{u}=\frac{1}{4e}\frac{e^{1+2x}}{1+2x}=\frac{e\cdot e^{2x}}{4e(1+2x)}=\frac{e^{2x}}{4+8x}+C. $$

Wolfram Alpha check

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  • $\begingroup$ Where did the $\frac{1}{4}$ come from? I see that you used the substitution rule, but am having trouble understanding all of your algebra $\endgroup$ – Evan Kim Feb 19 at 15:46
  • $\begingroup$ $$\frac{1}{4}(1+2x-1)\frac{d}{dx}(1+2x)=\frac{1}{4}2x\cdot2=\frac{1}{4}4x=x$$ $\endgroup$ – Michael Rybkin Feb 19 at 16:13

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