make a substitution to express the integrand as a rational function and then evaluate the integral how do i do this? problem inside!! thanks!!

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make a substitution to express the integrand as a rational function and then evaluate the integral how do i do this? problem inside!! thanks!!

Mathematics
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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=____ ?

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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 :(
Yes that's a good substitution! But when you have to take the derivative with respect to x that's why you have \[\frac{ du }{ dx }\]
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?
Yup, but note we have that +x in the denominator so you have solve for x here as well \[u=\sqrt{x+3}\] solve for x
\[\sqrt{x+3}=u,x+3=u^2,x=u^2-3,dx=2udu\]
\[du = \frac{ 1 }{ 2\sqrt{x+3} }dx\]
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!!
So our original integral is \[\int\limits \frac{ dx }{ 2\sqrt{x+3}+x }\] we made the substitution \[u= \sqrt{x+3} \implies du = \frac{ 1 }{ 2\sqrt{x+3} }dx\] but notice the integrand still has that +x in the denominator, and we can't have that when we want everything respect to u, so we look at our original substitution and see if there is a way we can get rid of the x, \[u=\sqrt{x+3} \implies x = u^2-3\] so we plug that in where that +x is, but note we can simplify our derivative \[du = \frac{ 1 }{ 2\sqrt{x+3} }dx \implies du = \frac{ 1 }{ 2u }dx \implies 2udu = dx\]
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
how can i do that? :/ sorry, i am having a hard time understanding calc :( i follow so far right now, but i do not understand what I am supposed to do when you say evaluate? is it finding the antiderivative?
Yes, when you integrate you find the antiderivative. You will have to use partial fractions
okay, i am not very good with partial fractions :( could you possibly guide me through the steps that I would do here? I just don't know what i do here :/ find A,B and C?
Well partial fractions can only be learnt through practice, you have to practice in calculus if you want to pass, so you should practice these problems after. Lets start of by looking at the integrand \[2 \int\limits \frac{ u }{ u^2+2u-3 }du\] lets factor the denominator what do you get?
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 :(
Ok so you will have to review factoring as well, you do this in high school, you should have got (u+3)(u-1)
So we have now \[2 \int\limits \frac{ u }{ (u+3)(u-1) }du\] this is our integral now
So for partial fractions we set it up as such \[\frac{ u }{ (u+3)(u-1) } = \frac{ A}{ (u+3) }+\frac{ B }{ (u-1) }\]
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 ?
Yeah, now we look at what we have on the left side, \[u=(A+B)+(-A+3B)\] now we can set up a system of equations \[A+B=1\] \[-A+3B=0\] I think you should be able to do the rest
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?
\[\frac{ u }{ (u+3)(u-1) } = \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }\] now this will become your integral so we have \[2 \int\limits \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }du\] this should be easy now, just evaluate
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
Noooo, all we have to do is go back and see \[u= \sqrt{x+3}\] \[\implies 2\left( \frac{ 3 }{ 4 }\ln|\sqrt{x+3}+3|+\frac{ 1 }{ 4 }\ln|\sqrt{x+3}-1| \right)+C\]
ohhh i see so what do we do next?
We're done, the problem not long enough? :P
hahaha ohhh okay :P yay!! thank you so ugh!! it was so long!! but i think i understand more of it now :) thanks so much!!! I'm having trouble understanding the concepts for calc right now so i appreciate your time and help!!
Np, integrals require lots of practice, I haven't mastered them myself xD, they can be tricky, they're not as easy as derivatives! So keep practicing!
:)

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