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

- anonymous

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

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- anonymous

Is that suppose to be 1/{(x+2)(x+3)(x+3)} What is squared?

- anonymous

no...it is \[1/({(x+2)\sqrt{x+3}})\]

- anonymous

how can this be solved?

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## More answers

- anonymous

What Calculus class are you in?

- anonymous

this can be solved using u subs?

- anonymous

What Calculus class are you in?

- anonymous

what?

- anonymous

Are you doing this for fun or a class?

- anonymous

class of course...wanted to learn

- anonymous

What type of math class? Clause it depends on what way.

- anonymous

i had learnt basic calculus...it is a bit advanced this time

- anonymous

"learned" so you're in basic calculus, trying to get ahead?

- anonymous

yeah

- anonymous

OK, what ways do you know how to use?

- anonymous

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..

- anonymous

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.

- anonymous

like denominator=(x+2).(x+3)\[^{1/2}\]

- anonymous

(x+2).(x+3)1/2

- anonymous

with the 1/2 on the (x+3) yes then pull it out so it becomes a product rule looking equation

- anonymous

then it would look like
\[1\div ((x+2).(x+3)^{1/2})\]

- anonymous

\[1\div \left( x+2 \right)\left( x+3 \right)^{1/2}\] to
\[\left( x+3 \right)^{-1/2} \] next to 1÷(x+2)

- anonymous

okay so now it is
\[\int\limits_{}^{} [(x+3)^{-1/2}]/(x+2)\]

- anonymous

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

- anonymous

Think product rule versus quot. rule

- anonymous

so can you solve this now?..i am solving it right now...plz do solve and let me know the answer..

- anonymous

This has taken longer than I planned, and I'm about to be sleeping on my keyboard. I hope this helps

- anonymous

correct me if i am wrong , but after intr. for the first time i get,
2\[\sqrt{x+3}/(x+2)-(2 \int\limits_{}^{}\sqrt{x+3} . \log(x+3))dx\]

- anonymous

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.

- anonymous

okay....i'll try by then....

- anonymous

did u get the solution?

- anonymous

yes, I'm going to type the whole solution out for you... hold on

- anonymous

thanks

- anonymous

\[\int\limits \frac{1}{(x+2)\sqrt {x+3}}dx \rightarrow u=\sqrt{x+3}, 2du=\frac{1}{\sqrt{x+3}}dx\]
\[2 \int\limits \frac{1}{x+2}du \rightarrow now:u^2-1=x+2, so \rightarrow 2 \int\limits \frac{1}{u^2-1}du\]
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:
\[since:u=\sqrt{x+3}\rightarrow \ln \left| \frac{\sqrt{x+3}-1}{\sqrt{x+3}+1}\right|+C\]

- anonymous

how did you take : u2−1=x+2 ?

- anonymous

like this:
\[u=\sqrt{x+3}\rightarrow u^2=(\sqrt{x+3})^2\rightarrow u^2=x+3\rightarrow u^2-1=x+2\]

- anonymous

ohh yeah..thanks nadeem

- anonymous

there's one more question...if u can answer...its regarding partial derivatives..can u?

- anonymous

sure

- anonymous

its just 2 places above this post.. which is like fxy =(-y sin xy)...so on...plz move to that post...

- anonymous

i got it :)

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