Can you solve $\int \frac{1}{2x+1} dx $ without substitution 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.
 A: 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 $$\int \frac{1}{2x+1} dx=\frac12\,\log|2x+1|+C.$$
A: 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.
A: 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 $x\ne -1/2$
