Is it possible to solve the integral $$\int \frac{1}{2x+1} dx $$ without substitution?
I can solve it with substitution, so don't worry. I just want to know if there is some nice trick without substitution.
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1$\begingroup$ $\frac{1}{2}\int\frac{2dx}{2x+1}=\frac{1}{2}\ln|2x+1|+C$ $\endgroup$– Juniven AcapulcoCommented Mar 16, 2017 at 0:36
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$\begingroup$ $\frac{1}{2x+1} = \frac{1}{2}\frac{1}{x+\frac{1}{2}}$ and $\frac{1}{x+\frac{1}{2}}$ is the derivative of $\log\left|x+\frac{1}{2}\right|$. $\endgroup$– StefanoCommented Mar 16, 2017 at 0:38
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$\begingroup$ If you consider reversing the chain-rule so you need to multiply the result by $\frac{1}{2}$ to "cancel" the $2$ produced by the chain rule and get $\frac{1}{2}\ln|2x+1| + c$. This is, however, just a perspective change on what u-substitution is. $\endgroup$– Dando18Commented Mar 16, 2017 at 0:39
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3 Answers
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Substitution is just a way to notice that something is the derivative of something else. In this case, with a bit of experience, one sees easily that $$\int \frac{1}{2x+1} dx=\frac12\,\log|2x+1|+C.$$
answered Mar 16, 2017 at 0:41
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$\begingroup$ Thank you for the answer. I was looking more for a series approach. $\endgroup$ Commented Mar 16, 2017 at 1:20
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No substitution required, just some geometric series:
$$\begin{align}\int\frac1{1+2x}\ dx&=\int1-2x+(2x)^2-(2x)^3+\dots\ dx\tag{$|x|<\frac12$}\\&=\int\sum_{n=0}^\infty(-1)^n2^nx^n\ dx\\&=\sum_{n=0}^\infty(-1)^n2^n\int x^n\ dx\\&=\sum_{n=0}^\infty(-1)^n2^n\frac{x^{n+1}}{n+1}+C\end{align}$$
Though probably not what you wanted? If you knew the geometric series to the natural logarithm, though, you'd see that
$$\sum_{n=0}^\infty(-1)^n2^n\frac{x^{n+1}}{n+1}=\frac12\sum_{n=1}^\infty(-1)^n\frac{(2x)^n}n=\frac12\ln(1+2x)$$
which holds by analytic continuation for $x\in\mathbb R$.
If you prefer, for $|x|>\frac12$, one could instead use
$$\frac1{1+2x}=\frac{\frac1{2x}}{1+\frac1{2x}}$$
and then applying the geometric series as Dr.MV has shown.
answered Mar 16, 2017 at 0:50
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1$\begingroup$ @Downvoter: I don't see what's wrong with this answer. Consider explaining why the downvote. $\endgroup$– Dando18Commented Mar 16, 2017 at 0:57
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$\begingroup$ Thank you for the answer. It is nice to see another approach. This was the stuff I was looking for. $\endgroup$ Commented Mar 16, 2017 at 1:19
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1$\begingroup$ I didn't down vote, but this isn't complete inasmuch as the series you wrote is not valid for $|x|>1/2$. I've posted a solution that accounts for this. ;-)) $\endgroup$ Commented Mar 16, 2017 at 1:23
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$\begingroup$ @Dr.MV :P That is true, nice catch. $\endgroup$ Commented Mar 16, 2017 at 1:45
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$\begingroup$ @WaldoRozir No problem :-) $\endgroup$ Commented Mar 16, 2017 at 1:45
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If you really wish to avoid simple substitution or any semblance thereof, we can write
$$\frac{1}{2x+1}=\sum_{n=0}^\infty (-1)^n(2x)^n$$
for $|x|< 1/2$. Then,
$$\begin{align} \int\frac{1}{2x+1}\,dx&=\frac12\sum_{n=1}^\infty (-1)^{n-1}\frac{(2x)^{n}}{n}+C\\\\ &=\frac{1}{2}\log(1+2x)+C \end{align}$$
If $|x|>1/2$, then we can write
$$\frac{1}{2x+1}=\sum_{n=0}^\infty(-1)^n(2x)^{-(n+1)}=\frac1{2x}+\frac12\sum_{n=1}^\infty(-1)^n(2x)^{-(n+1)}$$
and we have
$$\begin{align} \int\frac{1}{2x+1}\,dx&=\frac12\log(|x|)+\frac12\sum_{n=1}^\infty (-1)^{n-1}\frac{(2x)^{-n}}{n}+C\\\\ &=\frac{1}{2}\left(\log(|x|)+\log\left(1+\frac1{2x}\right)\right)+C\\\\ &=\frac12\log(|1+2x|)+C' \end{align}$$
Putting it all together, we have that
$$\int \frac1{2x+1}\,dx=\frac12\log(|1+2x|)+C$$
for all $x\ne -1/2$
answered Mar 16, 2017 at 0:57
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$\begingroup$ Thanks for the answer. This was what I searched for. $\endgroup$ Commented Mar 16, 2017 at 1:18
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2$\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$ Commented Mar 16, 2017 at 1:21