- anonymous

\[\int\limits_{}^{} (6x-11 / x+2) dx\]

- jamiebookeater

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- amistre64

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

- amistre64

frac{top}{bottom} makes a clean fraction

- anonymous

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

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## More answers

- anonymous

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

- amistre64

\[\frac{top}{australia}\]

- amistre64

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

- anonymous

thanks for the lesson in writing equations >:(

- anonymous

Do you get what I wrote msunprecedented?

- anonymous

no, i did not

- amistre64

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

- amistre64

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

- anonymous

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

- 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

- 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

- amistre64

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

- amistre64

and by left I mean right ;)

- amistre64

would the uv substitution work well there?

- 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)

- amistre64

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

- amistre64

##### 1 Attachment

- 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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- anonymous

I believe not.

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