anonymous
  • anonymous
find the integral using long division...?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{}^{}(27x^2)\div (3x+4)^3 dx\]
anonymous
  • anonymous
oh ok
myininaya
  • myininaya
Do we have to do long division? Can we use partial fractions instead?

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myininaya
  • myininaya
It actually says do long division?
anonymous
  • anonymous
Long division is going to get you nowhere in this case. Notice the power of the denominator is bigger than that of the numerator.
myininaya
  • myininaya
Oh that is true
myininaya
  • myininaya
good eye
anonymous
  • anonymous
ok, so how would you do it with partial fractions?
anonymous
  • anonymous
the 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
  • anonymous
Would you like a step by step?
anonymous
  • anonymous
if you could, that would help a lot!
anonymous
  • anonymous
No problem at all, first: how familiar are you with partial fractions?
anonymous
  • anonymous
i'm not familiar with the name but once i see it i will probably remember..
anonymous
  • anonymous
i don't think we have done partial fractions, i just looked it up on wolfram alpha and it doesn't look familiar
anonymous
  • anonymous
is there any other way to solve it?
anonymous
  • anonymous
Okay, 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
  • anonymous
ok now it looks familiar!!
anonymous
  • anonymous
so how would i get A B and C?
anonymous
  • anonymous
Okay, 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
  • anonymous
Is 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
  • anonymous
i took out constant first are my numbers correct?
anonymous
  • anonymous
Yeah, it would be just (constant)/27. So I believe so. Do you know how to integrate each of these terms?
anonymous
  • anonymous
well what would i use for my u value if i were to use u-substitution?
anonymous
  • anonymous
so that i only have to substitute once.
anonymous
  • anonymous
For all of them let u=3x+4; (1/3)du=dx
anonymous
  • anonymous
but i will still have to substitute again after
anonymous
  • anonymous
Not 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
  • anonymous
Thats 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 u-sub was.
anonymous
  • anonymous
i got \[(u^2\div2)+(8\div27u)-(8\div27u^2)+ c] is that correct?
anonymous
  • anonymous
\[(u^2/2)+(8/27u)-(8/27u^2)+c\]**
anonymous
  • anonymous
then plug in u-value?

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