Integrate by parts: $\int \ln (2x + 1) \, dx$

\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!

• 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 $\ln(2x+1)$. – Jared May 16 '13 at 21:04
• You're also missing two $\mathrm dx$, and parentheses around $\displaystyle1-\frac1{2x+1}$. – joriki May 16 '13 at 21:07
• And you can use substitution $t = 2x+1$ and then to apply integration by parts. – Cortizol May 16 '13 at 21:27

5 Answers

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

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

• 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} – seeker May 16 '13 at 21:12
• 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. – amWhy May 16 '13 at 21:21
• @amWhy: Nice to get ack'ed by OP! +1 – Amzoti May 17 '13 at 1:01

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 $$\left(x+\frac{1}{2}\right)\ln(2x+1) -x+C.$$

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,

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

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

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

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

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

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