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anonymous
 5 years ago
what is the integration of 1/{(x+2)(sqr of x+3)} ?
anonymous
 5 years ago
what is the integration of 1/{(x+2)(sqr of x+3)} ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that suppose to be 1/{(x+2)(x+3)(x+3)} What is squared?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no...it is \[1/({(x+2)\sqrt{x+3}})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how can this be solved?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What Calculus class are you in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this can be solved using u subs?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What Calculus class are you in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you doing this for fun or a class?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0class of course...wanted to learn

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What type of math class? Clause it depends on what way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i had learnt basic calculus...it is a bit advanced this time

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"learned" so you're in basic calculus, trying to get ahead?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, what ways do you know how to use?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like denominator=(x+2).(x+3)\[^{1/2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with the 1/2 on the (x+3) yes then pull it out so it becomes a product rule looking equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then it would look like \[1\div ((x+2).(x+3)^{1/2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1\div \left( x+2 \right)\left( x+3 \right)^{1/2}\] to \[\left( x+3 \right)^{1/2} \] next to 1÷(x+2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so now it is \[\int\limits_{}^{} [(x+3)^{1/2}]/(x+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Think product rule versus quot. rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so can you solve this now?..i am solving it right now...plz do solve and let me know the answer..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This has taken longer than I planned, and I'm about to be sleeping on my keyboard. I hope this helps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0correct me if i am wrong , but after intr. for the first time i get, 2\[\sqrt{x+3}/(x+2)(2 \int\limits_{}^{}\sqrt{x+3} . \log(x+3))dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay....i'll try by then....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did u get the solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, I'm going to type the whole solution out for you... hold on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{1}{(x+2)\sqrt {x+3}}dx \rightarrow u=\sqrt{x+3}, 2du=\frac{1}{\sqrt{x+3}}dx\] \[2 \int\limits \frac{1}{x+2}du \rightarrow now:u^21=x+2, so \rightarrow 2 \int\limits \frac{1}{u^21}du\] Use partial fraction decomposition: \[\frac{1}{u^21}=\frac {1}{(u+1)(u1)}= \frac{A}{u+1}+\frac {B}{u1}\] \[A(u1)+B(u+1)=1\rightarrow AuA+Bu+B=1\] \[(B+A)u=0u, BA=1\rightarrow so: B+A=0, BA=1\] solve for B and A: \[B=\frac{1}{2}, A= \frac{1}{2}\] \[2 \int\limits\limits \frac{1}{2} (\frac{1}{u1}\frac{1}{u+1})du \rightarrow \int\limits \frac{1}{u1}du \int\limits \frac{1}{u+1}du\] \[\int\limits\limits \frac{1}{u1}du \rightarrow v=u1, dv=du\] \[\rightarrow \int\limits\limits \frac{1}{v}dv=\ln(v)=\ln(u1)\] \[\int\limits \frac {1}{u+1}du \rightarrow z=u+1, dz=du\] \[\rightarrow \int\limits \frac{1}{z}dz=\ln(z)=\ln(u+1)\] so:\[\ln(u1)\ln(u+1)+C \rightarrow \ln \left \frac {u1}{u+1} \right+C\] From the begining we know what u is, so: \[since:u=\sqrt{x+3}\rightarrow \ln \left \frac{\sqrt{x+3}1}{\sqrt{x+3}+1}\right+C\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you take : u2−1=x+2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like this: \[u=\sqrt{x+3}\rightarrow u^2=(\sqrt{x+3})^2\rightarrow u^2=x+3\rightarrow u^21=x+2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh yeah..thanks nadeem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there's one more question...if u can answer...its regarding partial derivatives..can u?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its just 2 places above this post.. which is like fxy =(y sin xy)...so on...plz move to that post...
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