could someone look over my integration by substitution steps?

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could someone look over my integration by substitution steps?

Mathematics
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Question is to integrate \[\int\limits_{?}^{?} \frac{ 6x-1 }{ \sqrt{3x+1} }\]
i made u=3x+1 du/dx=3 so then dx=du/3
u=3x+1 \[x=\frac{ u-1 }{ 3 }\]

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Now I have \[\int\limits_{?}^{?} \frac{ 6(\frac{ u-1 }{ 3 }) }{ \sqrt{u} } \times \frac{ du }{ 3 }\]
Now I have \[\int\limits_{?}^{?} \frac{ 6(\frac{ u-1 }{ 3 })\color{red}{-1} }{ \sqrt{u} } \times \frac{ du }{ 3 }\]
then i did \[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ 2u-2 }{ \sqrt{u} }.du\]
then i did \[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ 2u-2 \color{red}{-1}}{ \sqrt{u} }.du\]
omg yes i forgot the plus one!!
omg yes i forgot the \(\color{Red}{minus}\) one!!
pellet i miss typed it.. lol 6x+1 .. Sorry!!
\[\int\limits_{?}^{?}\frac{ 6x+1 }{ \sqrt{3x+1} }\]
so now we are at \[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ 2u-1 }{ \sqrt{u} }\]
looks good, keep going..
then \[\frac{ 1 }{ 3 }\int\limits_{?}^{?} 2\sqrt{u} \times \frac{ 2 }{ \sqrt{u} }.du\]
then i integrated
do you mean \[\frac{ 1 }{ 3 }\int\limits_{?}^{?} 2\sqrt{u} - \frac{ 1 }{ \sqrt{u} }.du \]
\[\frac{ 1 }{ 3 } \times \frac{ 4u ^{\frac{ 3 }{ 2 }} }{ 3 }+4u ^{\frac{ 1 }{ 2 }}\]
oh gosh .....lol im a mess.. its a minus!
so then i think finally i did \[\frac{ 4u ^{\frac{ 3 }{ 2 }} }{ 9 }-\frac{ 4u ^{\frac{ 1 }{ 2 }} }{ 3 }\]
I think you should get \[\frac{ 1 }{ 3 } \left( \frac{ 4u ^{\frac{ 3 }{ 2 }} }{ 3 }-2u ^{\frac{ 1 }{ 2 }}\right)\]
yes cuz i did that error of not adding the plus one,..thanks so much! I really appreciate your time :D :D :D
np

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