Question

could someone look over my integration by substitution steps?

Answer 1

Question is to integrate
\[\int\limits_{?}^{?} \frac{ 6x-1 }{ \sqrt{3x+1} }\]

Answer 2

i made u=3x+1
du/dx=3 so then dx=du/3

Answer 3

u=3x+1
\[x=\frac{ u-1 }{ 3 }\]

Answer 4

Now I have
\[\int\limits_{?}^{?} \frac{ 6(\frac{ u-1 }{ 3 }) }{ \sqrt{u} } \times \frac{ du }{ 3 }\]

Answer 5

Now I have
\[\int\limits_{?}^{?} \frac{ 6(\frac{ u-1 }{ 3 })\color{red}{-1} }{ \sqrt{u} } \times \frac{ du }{ 3 }\]

Answer 6

then i did
\[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ 2u-2 }{ \sqrt{u} }.du\]

Answer 7

then i did
\[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ 2u-2 \color{red}{-1}}{ \sqrt{u} }.du\]

Answer 8

omg yes i forgot the plus one!!

Answer 9

omg yes i forgot the \(\color{Red}{minus}\) one!!

Answer 10

pellet i miss typed it.. lol
6x+1 .. Sorry!!

Answer 11

\[\int\limits_{?}^{?}\frac{ 6x+1 }{ \sqrt{3x+1} }\]

Answer 12

so now we are at \[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ 2u-1 }{ \sqrt{u} }\]

Answer 13

looks good, keep going..

Answer 14

then \[\frac{ 1 }{ 3 }\int\limits_{?}^{?} 2\sqrt{u} \times \frac{ 2 }{ \sqrt{u} }.du\]

Answer 15

then i integrated

Answer 16

do you mean
\[\frac{ 1 }{ 3 }\int\limits_{?}^{?} 2\sqrt{u} - \frac{ 1 }{ \sqrt{u} }.du \]

Answer 17

\[\frac{ 1 }{ 3 } \times \frac{ 4u ^{\frac{ 3 }{ 2 }} }{ 3 }+4u ^{\frac{ 1 }{ 2 }}\]

Answer 18

oh gosh .....lol im a mess..
its a minus!

Answer 19

so then i think finally i did
\[\frac{ 4u ^{\frac{ 3 }{ 2 }} }{ 9 }-\frac{ 4u ^{\frac{ 1 }{ 2 }} }{ 3 }\]

Answer 20

I think you should get
\[\frac{ 1 }{ 3 } \left( \frac{ 4u ^{\frac{ 3 }{ 2 }} }{ 3 }-2u ^{\frac{ 1 }{ 2 }}\right)\]

Answer 21

yes cuz i did that error of not adding the plus one,..thanks so much! I really appreciate your time :D :D :D

Answer 22

np