3
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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
4
$\begingroup$

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

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


NOTE:

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,

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

$\endgroup$
8
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

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.