Question

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

Answer 1

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

Answer 2

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

Answer 3

how can this be solved?

Answer 4

What Calculus class are you in?

Answer 5

this can be solved using u subs?

Answer 6

What Calculus class are you in?

Answer 7

what?

Answer 8

Are you doing this for fun or a class?

Answer 9

class of course...wanted to learn

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

yeah

Answer 14

OK, what ways do you know how to use?

Answer 15

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..

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

Think product rule versus quot. rule

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

did u get the solution?

Answer 31

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

Answer 32

thanks

Answer 33

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

Answer 34

how did you take : u2−1=x+2 ?

Answer 35

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

Answer 36

ohh yeah..thanks nadeem

Answer 37

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

Answer 38

sure

Answer 39

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

Answer 40

i got it :)