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

$u= 1+2x$

$du= 2 dx$

Now I saw a solution where a person took $xe^{2x}$ all together and used it as "du" but I did not understand how they were allowed to do that.

If "u" must consist of the first function of this mnemonic: L-I-A-T-E

L: Logarithmic Function

I: Inverse Function

A: Algebraic Function

T: Trigonometric Function

E: Exponential Function

Keep in mind I am only allowed to use Integration by Parts.

  • $\begingroup$ i really doubt that this integral is avaiable to the techniques u mention $\endgroup$
    – tired
    Dec 2 '15 at 16:08
  • $\begingroup$ Integration by parts works. Choose $\frac{1}{(1+2x)^2}$ as $du$ $\endgroup$
    – imranfat
    Dec 2 '15 at 16:09
  • $\begingroup$ @ imranfat why are you allowed to differentiate $xe^{2x}$?Isn't one an algebraic function, x and one and exponential function? $\endgroup$
    – Sunny
    Dec 2 '15 at 16:14
  • $\begingroup$ @Sunny Use the product rule to differentiate $xe^{2x}$. $\endgroup$
    – Mark Viola
    Dec 2 '15 at 16:15
  • 2
    $\begingroup$ Why is that relevant? $\endgroup$
    – Mark Viola
    Dec 2 '15 at 16:20

If one is restricted here as suggested in the OP, then we proceed as follows.

First, let $u=e^{2x}$ and $v=\frac{1}{4(2x+1)}+\frac14 \log(2x+1)$. So, $du=2e^{2x}$ and therefore we have

$$\begin{align} \int \frac{xe^{2x}}{(1+2x)^2}\,dx&=\frac{e^{2x}}{4(2x+1)}+\frac14 e^{2x}\log(2x+1)\\\\ &-\int \frac{e^{2x}}{2(2x+1)}\,dx-\frac12 \int e^{2x}\log(2x+1)\,dx \tag 1 \end{align}$$

Now, we integrate by part the first integral on the right-hand side if $(1)$ by setting $u=e^{2x}$ and $v=\frac14 \log(2x+1)$. So, $du=2e^{2x}\,dx$ and therefore we have

$$\begin{align} \int \frac{xe^{2x}}{(1+2x)^2}\,dx&=\frac{e^{2x}}{4(2x+1)}+\frac14 e^{2x}\log(2x+1)\\\\ &-\left(\frac14 e^{2x}\log(2x+1)-\frac12\int e^{2x}\log(2x+1)\,dx\right)-\frac12 \int e^{2x}\log(2x+1)\,dx\\\\ &=\frac{e^{2x}}{4(2x+1)}+C \end{align}$$


In using integration by parts, We are free to choose the $u$ and the $v$ functions. So, $u=xe^{2x}$ is perfectly acceptable and actually facilitates analysis. Let's see how to proceed.

If we let $u=xe^{2x}$, then $v=\frac{-1}{2(2x+1)}$. Proceeding, we have $du=(2x+1)e^{2x}\,dx$. Therefore,

$$\begin{align} \int \frac{xe^{2x}}{(1+2x)^2}\,dx&=\frac{xe^{2x}}{2(2x+1)}-\int \frac{-1}{2(2x+1)}(2x+1)e^{2x}\,dx\\\\ &=\frac{xe^{2x}}{2x+1}+\frac{e^{2x}}{4}+C\\\\ &=\frac{e^{2x}}{4(2x+1)}+C \end{align}$$

  • $\begingroup$ dv is $\frac{-1}{2(2x+1)}$? If so how are you calculating v? "v" would be $\frac{1}{(1+2x)^2}$ $\endgroup$
    – Sunny
    Dec 2 '15 at 16:43
  • $\begingroup$ Sunny. Are you talking about the second development - the one under "NOTE?" If so, $v=\frac{-1}{2(2x+1)}$, not $dv$. $dv=\frac{1}{(2x+1)^2}\,dx$ $\endgroup$
    – Mark Viola
    Dec 2 '15 at 16:45
  • $\begingroup$ @ Dr. MV Maybe it will be easier for me to follow if you tell me what dv is and then the antiderivative, v? $\endgroup$
    – Sunny
    Dec 2 '15 at 16:48
  • $\begingroup$ Sunny. You can find $dv$ by differentiating $v$. I suspect that that is something you can do on your own? $\endgroup$
    – Mark Viola
    Dec 2 '15 at 16:49
  • 1
    $\begingroup$ Integrate $1/(1+2x)^2$ to get $-1/(2(2x+1))$. $\endgroup$
    – Mark Viola
    Dec 2 '15 at 17:13

Consider the integral as presented by the proposer: $\int\frac{xe^{2x}}{(1+2x)^2}dx$ with $u = 1 + 2x$. This leads to \begin{align} \int\frac{xe^{2x}}{(1+2x)^2}dx &= \frac{1}{2} \, \int \left(\frac{u-1}{2} \right) \, e^{u-1} \, \frac{du}{u^{2}} \\ &= \frac{1}{4e} \, \int \left( \frac{e^{u}}{u} - \frac{e^{u}}{u^{2}} \right) \, du \\ &= \frac{1}{4e} \, \left[ Ei(u) - \left( Ei(u) - \frac{e^{u}}{u} \right) \right] + c_{0} \\ &= \frac{e^{u-1}}{4 \, u} + c_{0} \\ &= \frac{e^{2x}}{4 \, (1+2x)} + c_{0}. \end{align} Note that $Ei(x)$ is the Exponential integral.

This solution produces the same solution as that presented by Dr.MV

  • $\begingroup$ But you don't need $Ei(x)$ in this problem... $\endgroup$
    – imranfat
    Dec 2 '15 at 16:35
  • $\begingroup$ Interesting solution! $\endgroup$
    – Mark Viola
    Dec 2 '15 at 16:47

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