4
$\begingroup$

$$\eqalign{ & \int \ln (2x + 1) \, dx \cr & u = \ln (2x + 1) \cr & v = x \cr & {du \over dx} = {2 \over 2x + 1} \cr & {dv \over dx} = 1 \cr & \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr & = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr & = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr & = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr & = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$


The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$

Where did I go wrong?

Thanks!

$\endgroup$
  • 3
    $\begingroup$ Your last equality. You can't combine the log terms as you have done because only one of them has a factor of $x$. Instead, you should bring the $\frac{1}{2}$ down, and then factor out $\ln(2x+1)$. $\endgroup$ – Jared May 16 '13 at 21:04
  • $\begingroup$ You're also missing two $\mathrm dx$, and parentheses around $\displaystyle1-\frac1{2x+1}$. $\endgroup$ – joriki May 16 '13 at 21:07
  • $\begingroup$ And you can use substitution $t = 2x+1$ and then to apply integration by parts. $\endgroup$ – Cortizol May 16 '13 at 21:27
5
$\begingroup$

Starting from your second to last line (your integration was fine, minus a few $dx$'s in you integrals):

$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C \tag{1}$$

Good, up to this point... $\uparrow$.

So the error was in your last equality at the very end:

You made an error by ignoring the fact that the first term with $\ln(2x+1)$ as a factor also has $x$ as a factor, so we cannot multiply the arguments of $\ln$ to get $\ln(2x+1)^{3/2}$. What you could have done was first express $x\ln(2x+1) = \ln(2x+1)^x$ and then proceed as you did in your answer, but your result will then agree with your text's solution.

Alternatively, we can factor out like terms.

$$ = x\ln(2x + 1) + \frac 12 \ln(2x + 1) - x + C \tag{1}$$ $$= \color{blue}{\bf \frac 12 }{\cdot \bf 2x} \color{blue}{\bf \ln(2x+1)} + \color{blue}{\bf \frac 12 \ln(2x+1)}\cdot {\bf 1} - x + C$$

Factoring out $\color{blue}{\bf \frac 12 \ln(2x + 1)}$ gives us

$$= \left(\dfrac 12\ln(2x + 1)\right)\cdot \left(2x +1\right) - x + C $$ $$= \frac 12(2x + 1)\ln(2x+1) - x + C$$

$\endgroup$
  • $\begingroup$ Why can't I do: $$\eqalign{ & = \ln {(2x + 1)^x} + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C \cr & = \ln {(2x + 1)^{x + {1 \over 2}}} - x + C \cr} $$ $\endgroup$ – seeker May 16 '13 at 21:12
  • $\begingroup$ You're welcome, Assad: note, your procedure would have worked, had you done the arithmetic a little better: see the other answer, e.g. Note also that when it comes to integrals, especially when working with logarithms, many "different appearing solutions can be derived, depending on one's manipulations, etc. $\endgroup$ – amWhy May 16 '13 at 21:21
  • $\begingroup$ @amWhy: Nice to get ack'ed by OP! +1 $\endgroup$ – Amzoti May 17 '13 at 1:01
3
$\begingroup$

Here is a cute variant. Let $u=\ln(2x+1)$ and let $dv=dx$. Then $du=\frac{2}{2x+1}$ and (this is the cute part) we can take $v=x+\frac{1}{2}$. It follows that $$\int \ln(2x+1)\,dx=\left(x+\frac{1}{2}\right)\ln(2x+1)-\int \left(x+\frac{1}{2}\right)\frac{2}{2x+1}\,dx.$$ But the remaining integrand is just $1$! It follows that our integral is $$\left(x+\frac{1}{2}\right)\ln(2x+1) -x+C.$$

$\endgroup$
2
$\begingroup$

I am sure there are more tidy ways to do this but as an alternative...

Why not do $\int \ln(2x + 1)dx$ using:

$v^\prime = 1 \Rightarrow v = x$ and $u = \ln(2x+1)\Rightarrow u^\prime=\frac{2}{2x+1}$

Therefore,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{2x}{2x+1}dx$$

Then make the substitution $u=2x + 1$ to yield,

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

Thus,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}u - ln(u) + c$$

Then,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}(2x-1) - \ln(2x-1) + c$$

$$= \ln(2x-1)(x-1) - \frac{1}{2}(2x-1) + c$$

$\endgroup$
1
$\begingroup$

$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C $$

$$ = \ln(2x + 1)^x + \ln(2x + 1)^\dfrac 12 - x + C $$

$$= \ln({(2x + 1)^x \cdot(2x + 1)^\dfrac 12}) - x + C $$ $$= \ln{(2x + 1)^\dfrac{2x+1}{2} } - x + C $$ $$= \dfrac {1}{2}\cdot (2x+1) \ln{(2x + 1)} - x + C $$

$\endgroup$
1
$\begingroup$

Why don't you put $u = 2x + 1$ so that $du = 2 \,dx$? Then we'd have $$\int \log (2x + 1) \, dx = \frac {1} {2} \int \log u \, du = \frac {1} {2} (u \log u - u) = \frac {1} {2} (2x + 1) \log (2x + 1) - x - \frac {1} {2} + C.$$

$\endgroup$

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.