Q:

Why do I find two conflicting solutions when I calculate this integral $I = \int_{0}^{\infty} \frac{1}{{2x+1}} dx $?

I calculate this improper integral in two different ways but I find conflicting results. Can anybody identify why?
$$I = \int_{0}^{\infty} \frac{1}{{2x+1}} dx $$ 

 Solution 1 
$$ I = \frac{1}{2}\int_{0}^{\infty} \frac{(2x+1)'}{{2+1}} dx= \left[ \arctan(2x+1)\right]^{+\infty}_0 = \frac{3\pi}{4}  $$

 Solution 2 
$$\text{Let}\ u = 2x + 1$$
$$\text{Then}\ dx = \frac{1}{2} du $$
$$\text{and}\ \int_{0}^{\infty} \to \int_{1}^{\infty} $$
$$ I = \int_{1}^{\infty} \frac{1}{2u} du= \frac{1}{2}\left[ \ln(u)\right]^{+\infty}_1 = +\infty$$

A:

The solution 1 is wrong and the second one is correct. Indeed
$$ I = \frac{1}{2}\int_{0}^{\infty} \frac{(2x+1)'}{{2x+1}} dx=\frac{1}{2}[\log(2x+1)]_{0}^{\infty}=+\infty$$
On the other hand, note that
$$(\arctan(2x+1))'=\frac{(2x+1)'}{1+(2x+1)^2}=\frac{2}{1+(2x+1)^2}.$$

