INTEGRAL CALCULUS -Integral of logarithmic functions "Evaluate this integral of (x+3)dx/x+2

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INTEGRAL CALCULUS -Integral of logarithmic functions "Evaluate this integral of (x+3)dx/x+2

Mathematics
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I think you'll have to use long division first since both sides have the same power. The numerator's power must be < the denominator's power. :)
can u answer it? i want to compare my answer.. please?
alright, did you do the long division part?

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Other answers:

I got this after diving x+3 / x+2 : 1 + 1/x+2
dividing*
is this what you've got?
my answer x+ln(x+2)+c is this correct?
hold on , let me finish :)
yes, excellent :)
here another example can u answer this
sure, I'll give it a try ^_^
integral of xdx/x^2+4
you can expand the denominator and you'll get the following:\[x/(x^2+4) = x/(x+2)^2\] then you'll have to do the partial fractions part which is : \[x/(x+2)^2 = A/(x+2) + B/(x+2)^2\]....etc correct me if I'm wrong
you know how to integrate partial fractions, right? :)
wers the answer?
i dont know
lol, I have hinted the way for you for you to proceed ^_^
yes you know :) haven't you taken partial fractions?
oh wait
ther's a much simpler way to do it :) hold on no need for partial fractions here
ur answer? i want to compare my answer
\[\int\limits_{} x/(x^2+4) dx\] let : \[u = x^2 +4\] \[du = 2x dx\] so : \[= 1/2 \int\limits_{} 1/u du\] \[= 1/2(\ln |u|) + c\] \[=[ \ln|x^2+4|] /2 + c\] sorry abt the complicated mess earlier ^_^
is this what you've got?
1/2ln(X^2+4)+C
yes, well done ^_^
i wish you can transfer your knowledge in calculus to me
lol, I was so ignorant in it, but then worked hard to understand it. I took this in the beginning of the semester :), just be patient and practice more then you'll master it ^_^
do you have facebook i wish we can talk together to practice my enlish
english
Unfortunately, I don't have FB, myspace or any of these stuff since I don't have time for them :( , I'm sorry.
To perfect your english, read more, listen more and communicate more :) that's it
tnx
np :)

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