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\[\int\limits_{} \frac{ dx }{ 2\sqrt{x+3}+x }\]

What have you tried?

i am unsure of which concept i am using and how to apply
am i doing the u= ____ du=____ ?

Yes, when they ask for a substitution you can make u = something

okay, would it be u=2 (sqrt x+3) +x and du = dx ?

i am very confused :(

ohh okay, so what would i be using as substitution in this case then?

can i use this?
u=2 (sqrt x+3) +x and du = dx
?

You will be using \[u=\sqrt{x+3}\] what's the derivative of this?

would it be
1
-----
2 sqrt(x+3)
?

and if so, that would be du?

\[\sqrt{x+3}=u,x+3=u^2,x=u^2-3,dx=2udu\]

\[du = \frac{ 1 }{ 2\sqrt{x+3} }dx\]

ooh okie!! so what happens next? :/

do i still need to find more substitutions?

Well I asked you to solve for x, but surji already did it for you... now just plug it all in.

ohh okay, so i plug into the original equation?

Yes, that's why it's called substitution :)

like this? \[\int\limits_{} \frac{ 2udu }{ 2\sqrt{(u^2-3)+3}+(u^2-3) }\]

Nope, lets go over this slowly

aww:( okay!!

Now try and plug it all in

whoahh okay, let me try :)

\[\int\limits_{} \frac{ 2udu }{ 2u+(u^2-3) }\]

this?

Yes, exactly!

ooh okay yay!! so what do i do now?

Now you can evaluate it

Yes, when you integrate you find the antiderivative.
You will have to use partial fractions

i get (u+1)-1 ? :/

Factor the denominator u^2+2u-3

okay, so (u+1)(u-4) ? :/

(u+1)^2 + 1 ?

i don't really know :(

So we have now \[2 \int\limits \frac{ u }{ (u+3)(u-1) }du\] this is our integral now

ohhh okay, oops, sorry thought i had to do the thing where you add a 1 to the end or something :/

okay! and how do we solve for A and B ?

No you're thinking of completing the square, which is also helpful for some problems

ohh okay, oopsies :P

Multiply both sides by (u+3)(u-1) what do you get

A(u-1) and B(u+3) ?

Yes, but you must write it as \[u=A(u-1)+B(u+3)\]

Now distribute the A and B

ohh okay!! distribute like Au-A + Bu+3B ?

okay!! err, still a bit confused :/ what would i be doing next? do i sub in to the A/u+3 + B/u-1 ?

with A=1 and B=0? :/

This is just a regular system of equations solve for A and B

You can use any method, I recommend just adding as it's the most obvious in this case

ok, so we get A=3B --> 3B+B =1 =4B = 1 -->B=1/4
then A=3/4 ?

did i do that correctly?

Yeah, looks good!

yay! what do i do now?

so would i get this?\[\frac{ 1 }{ 2 } \log(1-u) + 3\log(u+3) + C \]

\[2(3/4\ln|u+3|+14\ln|u-1|)+C\] but you have to sub back the u substitution we made

1/4 not 14

okay! so we get this? \[2(3/4\ln \left| 5/4+3 \right|+ 1/4\ln \left| 5/4-1 \right| +C\]

5/4?

Curious, where did you get the 5/4 from?

i am not quite sure if i calculated correctly?
u=A+B -A +3B ?
=(3/4)+(1/4)-(3/4)+3(1/4)
=1 ? oops :P

ohhh i see
so what do we do next?

We're done, the problem not long enough? :P

:)