Need help checking a quick log integration problem

  • #1
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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)##
 
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Answers and Replies

  • #2
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Mathematica..
 

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