## marigirl one year ago could someone look over my integration by substitution steps?

1. marigirl

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

2. marigirl

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

3. marigirl

u=3x+1 $x=\frac{ u-1 }{ 3 }$

4. marigirl

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

5. ganeshie8

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

6. marigirl

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

7. ganeshie8

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

8. marigirl

omg yes i forgot the plus one!!

9. ganeshie8

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

10. marigirl

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

11. marigirl

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

12. marigirl

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

13. ganeshie8

looks good, keep going..

14. marigirl

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

15. marigirl

then i integrated

16. ganeshie8

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

17. marigirl

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

18. marigirl

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

19. marigirl

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

20. ganeshie8

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

21. marigirl

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

22. ganeshie8

np