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$$\int\ln(2x+1) \, dx$$

I setting up this problem and I am finding it hard to understand why $dv= 1$.

When using this formula $$\int u\ dv=uv-\int v\ du$$

And using these Guidelines for Selecting $u$ and $dv$:

“L-I-A-T-E” Choose $u$ to be the function that comes first in this list:

L: Logrithmic Function

I: Inverse Trig Function

A: Algebraic Function

T: Trig Function

E: Exponential Function

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  • $\begingroup$ Because your function can be written as $1*ln(2x+1)$ and so the $1$ will have to be chosen as your $dv$ which then becomes an $x$ as part of the term after the integral sign $\endgroup$
    – imranfat
    Nov 28, 2015 at 17:42
  • $\begingroup$ @imranfat What is one considered in the LIATE table? $\endgroup$
    – Sunny
    Nov 28, 2015 at 17:44
  • $\begingroup$ Do you have to use integration by parts? If not, it might be easier to use the substitution $y = 2x + 1$. $\endgroup$ Nov 28, 2015 at 17:47
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    $\begingroup$ Sunny. For the sake of some more practice of problems that falls exactly in the same category as yours, you can try to find the anti derivative of $arcsinx$, $arccosx$, and $arctanx$ in a similar fashion. You write $1$ in front of all of them and then choose $1=dv$. Good luck $\endgroup$
    – imranfat
    Nov 28, 2015 at 17:51
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    $\begingroup$ @Sunny I really agree with imranfat's previous comment. The arctrig functions make great practice integrals using this idea. $\endgroup$ Nov 28, 2015 at 17:54

4 Answers 4

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\begin{align} \int\underbrace{\ln(2x+1)}_\text{This is $u$.} \, dx & = \int u\,dx = ux - \int x\,du \\[10pt] & = x\ln(2x+1) - \int x\cdot \frac{2}{2x+1} \,dx \quad\text{etc.} \\[10pt] & = x\ln(2x+1) - \int \left( 1 - \frac 1 {2x+1} \right) \,dx \qquad \text{etc.} \end{align}

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  • $\begingroup$ Isn't the derivative of 2x+1 = 2? How did you get 2x on the second line? $\endgroup$
    – Sunny
    Nov 28, 2015 at 18:11
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    $\begingroup$ @Sunny The extra $x$ comes from the $v$ part of $v\,du$ $\endgroup$ Nov 28, 2015 at 18:17
  • $\begingroup$ Haste makes waste. I was right the first time. ${}\qquad{}$ $\endgroup$ Nov 28, 2015 at 21:10
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This is a really tricky problem to see for the first time, but it does fit with the guidelines you give.

The first choice if possible for $u$ is the logarithmic function. However, we also need another function in order to use integration by parts. What are we to do? Well, although it doesn't look like much, we can realize that $\ln(2x+1)=1\cdot\ln(2x+1)$, and now suddenly we have two functions and can try integration by parts. Turns out it works nicely enough too

$$\begin{align}\int\ln(2x+1)\,dx&=\int1\cdot\ln(2x+1)\,dx=x\ln(2x+1)-\int x\cdot\frac{2}{2x+1}\,dx \end{align}$$ and it's not too hard to finish off the integral from here.

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  • $\begingroup$ You just provided the entire answer...(sigh) $\endgroup$
    – imranfat
    Nov 28, 2015 at 17:49
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    $\begingroup$ @imranfat I didn't think that the OP had a problem with getting the answer, just understanding how to set up the problem, but it's easy enough for me to get rid of the last few steps $\endgroup$ Nov 28, 2015 at 17:51
  • $\begingroup$ Yeah, the set up was her issue I think. I should be straight forward now $\endgroup$
    – imranfat
    Nov 28, 2015 at 17:53
  • $\begingroup$ at this point I am $stuck\ \\lnx(2x+1)-\int\frac{1}{2x+1}dx$ I u subbed 2x+1 and my 2 canceled out. However, in your answer this is not the case. $\endgroup$
    – Sunny
    Nov 28, 2015 at 18:01
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    $\begingroup$ @Sunny. What do you mean the $2$ canceled? If you sub $2x+1=u$ you get $2dx=du$ or $dx=1/2du$ You new integral is of the form $\frac{1}{2}lnudu$ which can be written as $1*lnudu$ $\endgroup$
    – imranfat
    Nov 28, 2015 at 18:07
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$$\int\ln(2x+1)\ \mathrm dx$$ Using integration by parts, we have $$u=\ln(2x+1)\Rightarrow \mathrm du=\frac{2}{2x+1}\mathrm dx$$ $$\mathrm dv=\mathrm dx\Rightarrow v=x$$ Which yields $$x\ln(2x+1)-\int\frac{2x}{2x+1}\mathrm dx$$ Using substitution, we have $$s=2x+1\Rightarrow \frac12\mathrm ds=\mathrm dx$$ Therefore $$x\ln(2x+1)-\frac12\int\frac{s-1}{s}\mathrm ds$$ $$=x\ln(2x+1)-\frac12\int\left(1-\frac{1}{s}\right)\ \mathrm ds$$ $$=x\ln(2x+1)-\frac12\left(\int\mathrm ds-\int\frac{1}{s}\mathrm ds\right)$$ $$=x\ln(2x+1)-\frac12\int\mathrm ds+\frac12\int\frac{1}{s}\mathrm ds$$ $$=x\ln(2x+1)-\frac12 s+\frac12\ln s+C$$ $$=x\ln(2x+1)-\frac12(2x+1)+\frac12\ln(2x+1)+C$$ $$=\left(x+\frac12\right)\ln(2x+1)-\left(x+\frac12\right)+C$$ $$=\left(x+\frac12\right)\big(\ln(2x+1)-1\big)+C$$

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  • $\begingroup$ Your using $s$ twice. I guess I have never seen this. $s=2x+1$ and $s-1=2x$ What does one $call$ what you have done? And when do you $use$ this method? $\endgroup$
    – Sunny
    Nov 28, 2015 at 18:57
  • $\begingroup$ Just for confirmation: When you break up an integral$\int f(x)+g(x)=\int f(x)+\int g(x)$ You place the constant on both integrals $\frac{1}{2}\int f(x)+g(x)=$ $\frac{1}{2}\int f(x)+\frac{1}{2}\int g(x)$ $\endgroup$
    – Sunny
    Nov 28, 2015 at 19:02
  • $\begingroup$ k170 can you answer my questions? $\endgroup$
    – Sunny
    Nov 28, 2015 at 19:06
  • $\begingroup$ @Sunny, the method I used is called integration by substitution. And the answer to your last comment is yes. $\endgroup$
    – k170
    Nov 28, 2015 at 19:07
  • $\begingroup$ I guess this is using u substitution in several or many ways. Ok thanks. How about my second question? $\endgroup$
    – Sunny
    Nov 28, 2015 at 19:10
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Another solution by substitution:

$u = 2x + 1$

$x = \frac{u-1}{2}$

$\frac{du}{dx} = \frac{1}{2}$

$\int \log(2x+1) dx = \frac{1}{2}\int \log(u) du = \frac{1}{2}(u \log(u) - u) + C = \frac{1}{2}((2x+1) log(2x+1) - (2x+1)) + C$

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