## anonymous 5 years ago $\int\limits_{}^{} (6x-11 / x+2) dx$

1. amistre64

nice job with the equation writing skill...kudos ;)

2. amistre64

frac{top}{bottom} makes a clean fraction

3. anonymous

You need to substitute x for x=u+2 and then solve (dx=du since d/dy of +2 is zero).

4. anonymous

Err... that's not d/dy of +2 but the derivative of +2

5. amistre64

$\frac{top}{australia}$

6. amistre64

$\int \frac{6x-11}{x+2}dx$

7. anonymous

thanks for the lesson in writing equations >:(

8. anonymous

Do you get what I wrote msunprecedented?

9. anonymous

no, i did not

10. amistre64

lol.... I was impressed by your talent ...still am :)

11. amistre64

do you want the integral of 3 seperate terms; or is what I posted a correct interpretation of it?

12. anonymous

no its correct $\int\limits_{}^{} \frac {6x-11} {x+2} dx$

13. amistre64

... factoring might be possible but we would have to find out what value had been canceled before hand.... and then do partial decomposition as a possibility

14. anonymous

Actually, this is really simple if you view the fraction like this: $6x/(x+2) - (11/x+2)$ Now that it's in this form, you split the integral into two : $\int\limits_{?}^{?}6x/(x+2)dx - \int\limits_{?}^{?}11/(6x+2)dx$ This should be easier to integrate... let me know if you need more help

15. amistre64

the left is easy now :) the right still has a straggler.... that "x" up top can get bothersome

16. amistre64

and by left I mean right ;)

17. amistre64

would the uv substitution work well there?

18. anonymous

uv substitution wouldn't work because I don't think you could factor x+2. (I can't see a way to factor anyway)

19. amistre64

6x goes to zero pretty quick like; but I get ln(x+2) and get lost on integrateing that

20. amistre64

21. anonymous

Hmm.. I think integration by substitution works here. If you let u=x+2, you can get the right equal to:$\int\limits_{?}^{?} (6u-12)/u *du$ Then you split the fraction one more time

22. anonymous

I mean the left when I say the "right" above

23. amistre64

lol..... stage left ...STAGE left!! :)

24. anonymous

Yikes! I'm soo confused! U-sub here? Integration by parts there? :(

25. anonymous

msunprecedented, have you learnt how to do integration by substitution yet?

26. anonymous

I believe not.