Need help checking a quick log integration problem

[Eclair_de_XII]

Homework Statement

"Find $\int_0^x \frac{6\ dt}{2t+1}$."

Homework Equations

$\int \frac{\ dx}{x} = \ ln(x)$ for $x>0$

The Attempt at a Solution

Method 1: Let $u=2t+1$, $\frac{1}{2} du=\ dt$.
$\int_0^x \frac{6\ dt}{2t+1}=3\int_1^{2t+1} \frac{du}{u}=3\ ln(u)|_1^{2t+1}=3\ ln(2t+1)$

Method 2: $\int_0^x \frac{6\ dt}{2t+1}=\frac{0.5}{0.5} \int_0^x \frac{6\ dt}{2t+1}=\int_0^x \frac{3\ dt}{t+\frac{1}{2}}=3\ ln(t+\frac{1}{2})|_0^x=3[\ ln(x+\frac{1}{2})-\ ln(\frac{1}{2})]=3\ ln(\frac{x+\frac{1}{2}}{\frac{1}{2}})$

My book says that method 1 gives the correct answer, but I cannot figure out why method 2 does not.

Edit: Wait, never mind; I think I understand...

$3\ ln(\frac{x+\frac{1}{2}}{\frac{1}{2}})=3\ ln(\frac{2}{2}\frac{x+\frac{1}{2}}{\frac{1}{2}})=3\ ln(\frac{2x+1}{1})=3\ ln(2x+1)-3\ln(1)=3\ ln(2x+1)$

 

[MidgetDwarf]

Mathematica..