anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{} \frac{ dx }{ 2\sqrt{x+3}+x }\]
Astrophysics
  • Astrophysics
What have you tried?
anonymous
  • anonymous
i am unsure of which concept i am using and how to apply am i doing the u= ____ du=____ ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Astrophysics
  • Astrophysics
Yes, when they ask for a substitution you can make u = something
anonymous
  • anonymous
okay, would it be u=2 (sqrt x+3) +x and du = dx ?
anonymous
  • anonymous
i am very confused :(
Astrophysics
  • 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
  • anonymous
ohh okay, so what would i be using as substitution in this case then?
anonymous
  • anonymous
can i use this? u=2 (sqrt x+3) +x and du = dx ?
Astrophysics
  • Astrophysics
You will be using \[u=\sqrt{x+3}\] what's the derivative of this?
anonymous
  • anonymous
would it be 1 ----- 2 sqrt(x+3) ?
anonymous
  • anonymous
and if so, that would be du?
Astrophysics
  • 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
  • anonymous
\[\sqrt{x+3}=u,x+3=u^2,x=u^2-3,dx=2udu\]
Astrophysics
  • Astrophysics
\[du = \frac{ 1 }{ 2\sqrt{x+3} }dx\]
anonymous
  • anonymous
ooh okie!! so what happens next? :/
anonymous
  • anonymous
do i still need to find more substitutions?
Astrophysics
  • Astrophysics
Well I asked you to solve for x, but surji already did it for you... now just plug it all in.
anonymous
  • anonymous
ohh okay, so i plug into the original equation?
Astrophysics
  • Astrophysics
Yes, that's why it's called substitution :)
anonymous
  • anonymous
like this? \[\int\limits_{} \frac{ 2udu }{ 2\sqrt{(u^2-3)+3}+(u^2-3) }\]
Astrophysics
  • Astrophysics
Nope, lets go over this slowly
anonymous
  • anonymous
aww:( okay!!
Astrophysics
  • 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
  • Astrophysics
Now try and plug it all in
anonymous
  • anonymous
whoahh okay, let me try :)
anonymous
  • anonymous
\[\int\limits_{} \frac{ 2udu }{ 2u+(u^2-3) }\]
anonymous
  • anonymous
this?
Astrophysics
  • Astrophysics
Yes, exactly!
anonymous
  • anonymous
ooh okay yay!! so what do i do now?
Astrophysics
  • Astrophysics
Now you can evaluate it
anonymous
  • 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
  • Astrophysics
Yes, when you integrate you find the antiderivative. You will have to use partial fractions
anonymous
  • 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
  • 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
  • anonymous
i get (u+1)-1 ? :/
Astrophysics
  • Astrophysics
Factor the denominator u^2+2u-3
anonymous
  • anonymous
okay, so (u+1)(u-4) ? :/
anonymous
  • anonymous
(u+1)^2 + 1 ?
anonymous
  • anonymous
i don't really know :(
Astrophysics
  • 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
  • Astrophysics
So we have now \[2 \int\limits \frac{ u }{ (u+3)(u-1) }du\] this is our integral now
Astrophysics
  • 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
  • anonymous
ohhh okay, oops, sorry thought i had to do the thing where you add a 1 to the end or something :/
anonymous
  • anonymous
okay! and how do we solve for A and B ?
Astrophysics
  • Astrophysics
No you're thinking of completing the square, which is also helpful for some problems
anonymous
  • anonymous
ohh okay, oopsies :P
Astrophysics
  • Astrophysics
Multiply both sides by (u+3)(u-1) what do you get
anonymous
  • anonymous
A(u-1) and B(u+3) ?
Astrophysics
  • Astrophysics
Yes, but you must write it as \[u=A(u-1)+B(u+3)\]
Astrophysics
  • Astrophysics
Now distribute the A and B
anonymous
  • anonymous
ohh okay!! distribute like Au-A + Bu+3B ?
Astrophysics
  • 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
  • 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
  • anonymous
with A=1 and B=0? :/
Astrophysics
  • Astrophysics
This is just a regular system of equations solve for A and B
Astrophysics
  • Astrophysics
You can use any method, I recommend just adding as it's the most obvious in this case
anonymous
  • anonymous
ok, so we get A=3B --> 3B+B =1 =4B = 1 -->B=1/4 then A=3/4 ?
anonymous
  • anonymous
did i do that correctly?
Astrophysics
  • Astrophysics
Yeah, looks good!
anonymous
  • anonymous
yay! what do i do now?
Astrophysics
  • 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
  • anonymous
so would i get this?\[\frac{ 1 }{ 2 } \log(1-u) + 3\log(u+3) + C \]
Astrophysics
  • Astrophysics
\[2(3/4\ln|u+3|+14\ln|u-1|)+C\] but you have to sub back the u substitution we made
Astrophysics
  • Astrophysics
1/4 not 14
anonymous
  • 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
  • Astrophysics
5/4?
Astrophysics
  • Astrophysics
Curious, where did you get the 5/4 from?
anonymous
  • 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
  • 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
  • anonymous
ohhh i see so what do we do next?
Astrophysics
  • Astrophysics
We're done, the problem not long enough? :P
anonymous
  • 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
  • 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
  • anonymous
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.