what is the integration of 1/{(x+2)(sqr of x+3)} ?
Is that suppose to be 1/{(x+2)(x+3)(x+3)} What is squared?
no...it is \[1/({(x+2)\sqrt{x+3}})\]
how can this be solved?
What Calculus class are you in?
this can be solved using u subs?
What Calculus class are you in?
what?
Are you doing this for fun or a class?
class of course...wanted to learn
What type of math class? Clause it depends on what way.
i had learnt basic calculus...it is a bit advanced this time
"learned" so you're in basic calculus, trying to get ahead?
yeah
OK, what ways do you know how to use?
for this prob, i guess we can use integration by parts..right? but i always get confused in which term to take as 'u' and which as 'v'....i need some hints..
In order to do that your going to need to change some parts and move them around a little. I would change the square root to a power and take it out from the demomatator.
like denominator=(x+2).(x+3)\[^{1/2}\]
(x+2).(x+3)1/2
with the 1/2 on the (x+3) yes then pull it out so it becomes a product rule looking equation
then it would look like \[1\div ((x+2).(x+3)^{1/2})\]
\[1\div \left( x+2 \right)\left( x+3 \right)^{1/2}\] to \[\left( x+3 \right)^{-1/2} \] next to 1÷(x+2)
okay so now it is \[\int\limits_{}^{} [(x+3)^{-1/2}]/(x+2)\]
you could look at it that way but it might be more helpful if you had that in two separate parts. might be easier to work with
Think product rule versus quot. rule
so can you solve this now?..i am solving it right now...plz do solve and let me know the answer..
This has taken longer than I planned, and I'm about to be sleeping on my keyboard. I hope this helps
correct me if i am wrong , but after intr. for the first time i get, 2\[\sqrt{x+3}/(x+2)-(2 \int\limits_{}^{}\sqrt{x+3} . \log(x+3))dx\]
I think your missing a /2 in the first part, but my brain is turning off. night, sorry but I'm just too tired to think much more, worst case I'll check back tomorrow.
okay....i'll try by then....
did u get the solution?
yes, I'm going to type the whole solution out for you... hold on
thanks
\[\int\limits \frac{1}{(x+2)\sqrt {x+3}}dx \rightarrow u=\sqrt{x+3}, 2du=\frac{1}{\sqrt{x+3}}dx\] \[2 \int\limits \frac{1}{x+2}du \rightarrow now:u^2-1=x+2, so \rightarrow 2 \int\limits \frac{1}{u^2-1}du\] Use partial fraction decomposition: \[\frac{1}{u^2-1}=\frac {1}{(u+1)(u-1)}= \frac{A}{u+1}+\frac {B}{u-1}\] \[A(u-1)+B(u+1)=1\rightarrow Au-A+Bu+B=1\] \[(B+A)u=0u, B-A=1\rightarrow so: B+A=0, B-A=1\] solve for B and A: \[B=\frac{1}{2}, A=- \frac{1}{2}\] \[2 \int\limits\limits \frac{1}{2} (\frac{1}{u-1}-\frac{1}{u+1})du \rightarrow \int\limits \frac{1}{u-1}du- \int\limits \frac{1}{u+1}du\] \[\int\limits\limits \frac{1}{u-1}du \rightarrow v=u-1, dv=du\] \[\rightarrow \int\limits\limits \frac{1}{v}dv=\ln(v)=\ln(u-1)\] \[\int\limits \frac {1}{u+1}du \rightarrow z=u+1, dz=du\] \[\rightarrow \int\limits \frac{1}{z}dz=\ln(z)=\ln(u+1)\] so:\[\ln(u-1)-\ln(u+1)+C \rightarrow \ln \left| \frac {u-1}{u+1} \right|+C\] From the begining we know what u is, so: \[since:u=\sqrt{x+3}\rightarrow \ln \left| \frac{\sqrt{x+3}-1}{\sqrt{x+3}+1}\right|+C\]
how did you take : u2−1=x+2 ?
like this: \[u=\sqrt{x+3}\rightarrow u^2=(\sqrt{x+3})^2\rightarrow u^2=x+3\rightarrow u^2-1=x+2\]
ohh yeah..thanks nadeem
there's one more question...if u can answer...its regarding partial derivatives..can u?
sure
its just 2 places above this post.. which is like fxy =(-y sin xy)...so on...plz move to that post...
i got it :)
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