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anonymous
 one year ago
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
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} \frac{ dx }{ 2\sqrt{x+3}+x }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2What have you tried?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am unsure of which concept i am using and how to apply am i doing the u= ____ du=____ ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yes, when they ask for a substitution you can make u = something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, would it be u=2 (sqrt x+3) +x and du = dx ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am very confused :(

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay, so what would i be using as substitution in this case then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can i use this? u=2 (sqrt x+3) +x and du = dx ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2You will be using \[u=\sqrt{x+3}\] what's the derivative of this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be 1  2 sqrt(x+3) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and if so, that would be du?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yup, 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x+3}=u,x+3=u^2,x=u^23,dx=2udu\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[du = \frac{ 1 }{ 2\sqrt{x+3} }dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh okie!! so what happens next? :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i still need to find more substitutions?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Well I asked you to solve for x, but surji already did it for you... now just plug it all in.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay, so i plug into the original equation?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yes, that's why it's called substitution :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like this? \[\int\limits_{} \frac{ 2udu }{ 2\sqrt{(u^23)+3}+(u^23) }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Nope, lets go over this slowly

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So 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, \[u=\sqrt{x+3} \implies x = u^23\] 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
 one year ago
Best ResponseYou've already chosen the best response.2Now try and plug it all in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whoahh okay, let me try :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} \frac{ 2udu }{ 2u+(u^23) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh okay yay!! so what do i do now?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Now you can evaluate it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how 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
 one year ago
Best ResponseYou've already chosen the best response.2Yes, when you integrate you find the antiderivative. You will have to use partial fractions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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
 one year ago
Best ResponseYou've already chosen the best response.2Well 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 \[2 \int\limits \frac{ u }{ u^2+2u3 }du\] lets factor the denominator what do you get?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Factor the denominator u^2+2u3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, so (u+1)(u4) ? :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't really know :(

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Ok so you will have to review factoring as well, you do this in high school, you should have got (u+3)(u1)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So we have now \[2 \int\limits \frac{ u }{ (u+3)(u1) }du\] this is our integral now

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So for partial fractions we set it up as such \[\frac{ u }{ (u+3)(u1) } = \frac{ A}{ (u+3) }+\frac{ B }{ (u1) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh okay, oops, sorry thought i had to do the thing where you add a 1 to the end or something :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! and how do we solve for A and B ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2No you're thinking of completing the square, which is also helpful for some problems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay, oopsies :P

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Multiply both sides by (u+3)(u1) what do you get

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yes, but you must write it as \[u=A(u1)+B(u+3)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Now distribute the A and B

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay!! distribute like AuA + Bu+3B ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.0okay!! err, still a bit confused :/ what would i be doing next? do i sub in to the A/u+3 + B/u1 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with A=1 and B=0? :/

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2This is just a regular system of equations solve for A and B

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2You can use any method, I recommend just adding as it's the most obvious in this case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so we get A=3B > 3B+B =1 =4B = 1 >B=1/4 then A=3/4 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did i do that correctly?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, looks good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay! what do i do now?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ u }{ (u+3)(u1) } = \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u1) }\] now this will become your integral so we have \[2 \int\limits \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u1) }du\] this should be easy now, just evaluate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would i get this?\[\frac{ 1 }{ 2 } \log(1u) + 3\log(u+3) + C \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[2(3/4\lnu+3+14\lnu1)+C\] but you have to sub back the u substitution we made

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! so we get this? \[2(3/4\ln \left 5/4+3 \right+ 1/4\ln \left 5/41 \right +C\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Curious, where did you get the 5/4 from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.2Noooo, 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
 one year ago
Best ResponseYou've already chosen the best response.0ohhh i see so what do we do next?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2We're done, the problem not long enough? :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahaha 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
 one year ago
Best ResponseYou've already chosen the best response.2Np, 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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