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anonymous
 4 years ago
find the integral using long division...?
anonymous
 4 years ago
find the integral using long division...?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}(27x^2)\div (3x+4)^3 dx\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1Do we have to do long division? Can we use partial fractions instead?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1It actually says do long division?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Long division is going to get you nowhere in this case. Notice the power of the denominator is bigger than that of the numerator.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, so how would you do it with partial fractions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the other parts of the question i had to use long division so i figured it would be the same throughout.. guess thats why i got confused.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Would you like a step by step?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you could, that would help a lot!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem at all, first: how familiar are you with partial fractions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm not familiar with the name but once i see it i will probably remember..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't think we have done partial fractions, i just looked it up on wolfram alpha and it doesn't look familiar

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is there any other way to solve it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, well in general, partial fractions is a way to take rational expression: \[\frac{P(x)}{Q(x)}\] And write it as a sum of lower order terms. Think of it as "undoing" a common denominator. So we have: \[\frac{27x^2}{(3x+4)^3}=\frac{A}{3x+4}+\frac{B}{(3x+4)^2}+\frac{C}{(3x+4)^3} \implies \] \[A(3x+4)^2+B(3x+4)+C=27x^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok now it looks familiar!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how would i get A B and C?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, to do that we simply multiply it all out, and set the coefficients of the terms equal. Sounds odd but let me show you what I mean. First, you notice that if you let x=4/3 that the 3x+4 terms go to zero? So lets do that, let x=4/3 then we get: \[A(0)^2+B(0)+C=27(4/3)^2 \implies C=27(16/9)=3*16=48\] Now we know C! So we have: \[A(3x+4)^2+(3x+4)B+48=27x^2\] Now we can do my method, so we have: \[A(9x^2+16+24x)+B(3x+4)+48=27x^2 \] Now we can deduce what we want to know. Looking at this, you realize for the sides to be equal, the coefficient 27 must equal the coefficients in front of the x^2 terms on the left. So for this example: \[27=9A \implies A=3; 24A+3B=0, A=3 \implies B=24\] Thus we have: \[\int\limits \frac{27x^2}{(3x+4)^3}dx=\int\limits \left[ \frac{3}{3x+4}\frac{24}{(3x+4)^2}+\frac{48}{(3x+4)^3} \right]dx\] Follow me to here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is it ok if i wrote it:\[27\int\limits_{}^{}1\div(9(3x+4)8\div9(3x+4)^2+16\div9(3x+4)^3 dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i took out constant first are my numbers correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, it would be just (constant)/27. So I believe so. Do you know how to integrate each of these terms?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well what would i use for my u value if i were to use usubstitution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so that i only have to substitute once.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For all of them let u=3x+4; (1/3)du=dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but i will still have to substitute again after

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not quite: \[9 \int\limits \frac{du}{9u}8 \int\limits \frac{du}{27u^2}+16 \int\limits \frac{du}{27u^3}\] See if you agree. Note: I used your numbers so the constants may be slightly off of what I have above^^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thats doing all of it in 1 substitution. Obviously when you integrate it you'll have a function of u, then you replug in what your usub was.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got \[(u^2\div2)+(8\div27u)(8\div27u^2)+ c] is that correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(u^2/2)+(8/27u)(8/27u^2)+c\]**

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then plug in uvalue?
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