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

Question

I calculate this improper integral in two different ways but I find conflicting results. Can anybody identify why?

$$I = \int_{0}^{\infty} \frac{1}{{2x+1}} dx $$

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Solution 1

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

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Solution 2

$$\text{Let}\ u = 2x + 1$$

$$\text{Then}\ dx = \frac{1}{2} du $$ $$\text{and}\ \int_{0}^{\infty} \to \int_{1}^{\infty} $$

$$ I = \int_{1}^{\infty} \frac{1}{2u} du= \frac{1}{2}\left[ \ln(u)\right]^{+\infty}_1 = +\infty$$

Answer 1

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 $$(\arctan(2x+1))'=\frac{(2x+1)'}{1+(2x+1)^2}=\frac{2}{1+(2x+1)^2}.$$

Answer 2

Integration by substitution:

1) $\phi(x):=2x+1$; $f(x)=\frac{1}{x};$

$(1/2)\int_{0}^{\infty}f(\phi(t))\phi'(t)dt=$

$(1/2)\int_{\phi(0)}^{\phi(\infty)}f(x)dx=$

$(1/2)\int_{1}^{\infty}\frac{1}{x}dx=(1/2)\log x]_1^{\infty}$.