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

- anonymous

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- anonymous

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

- Astrophysics

What have you tried?

- anonymous

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

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## More answers

- Astrophysics

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

- anonymous

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

- anonymous

i am very confused :(

- Astrophysics

Yes that's a good substitution! But when you have to take the derivative with respect to x that's why you have \[\frac{ du }{ dx }\]

- anonymous

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

- anonymous

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

- Astrophysics

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

- anonymous

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

- anonymous

and if so, that would be du?

- Astrophysics

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

- anonymous

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

- Astrophysics

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

- anonymous

ooh okie!! so what happens next? :/

- anonymous

do i still need to find more substitutions?

- Astrophysics

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

- anonymous

ohh okay, so i plug into the original equation?

- Astrophysics

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

- anonymous

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

- Astrophysics

Nope, lets go over this slowly

- anonymous

aww:( okay!!

- Astrophysics

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

- Astrophysics

Now try and plug it all in

- anonymous

whoahh okay, let me try :)

- anonymous

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

- anonymous

this?

- Astrophysics

Yes, exactly!

- anonymous

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

- Astrophysics

Now you can evaluate it

- anonymous

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?

- Astrophysics

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

- anonymous

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?

- Astrophysics

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?

- anonymous

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

- Astrophysics

Factor the denominator u^2+2u-3

- anonymous

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

- anonymous

(u+1)^2 + 1 ?

- anonymous

i don't really know :(

- Astrophysics

Ok so you will have to review factoring as well, you do this in high school, you should have got (u+3)(u-1)

- Astrophysics

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

- Astrophysics

So for partial fractions we set it up as such \[\frac{ u }{ (u+3)(u-1) } = \frac{ A}{ (u+3) }+\frac{ B }{ (u-1) }\]

- anonymous

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

- anonymous

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

- Astrophysics

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

- anonymous

ohh okay, oopsies :P

- Astrophysics

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

- anonymous

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

- Astrophysics

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

- Astrophysics

Now distribute the A and B

- anonymous

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

- Astrophysics

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

- anonymous

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

- anonymous

with A=1 and B=0? :/

- Astrophysics

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

- Astrophysics

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

- anonymous

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

- anonymous

did i do that correctly?

- Astrophysics

Yeah, looks good!

- anonymous

yay! what do i do now?

- Astrophysics

\[\frac{ u }{ (u+3)(u-1) } = \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }\] 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

- anonymous

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

- Astrophysics

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

- Astrophysics

1/4 not 14

- anonymous

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

- Astrophysics

5/4?

- Astrophysics

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

- anonymous

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

- Astrophysics

Noooo, all we have to do is go back and see \[u= \sqrt{x+3}\] \[\implies 2\left( \frac{ 3 }{ 4 }\ln|\sqrt{x+3}+3|+\frac{ 1 }{ 4 }\ln|\sqrt{x+3}-1| \right)+C\]

- anonymous

ohhh i see
so what do we do next?

- Astrophysics

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

- anonymous

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!!

- Astrophysics

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!

- anonymous

:)

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