find the integral using long division...?

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find the integral using long division...?

Mathematics
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\[\int\limits_{}^{}(27x^2)\div (3x+4)^3 dx\]
oh ok
Do we have to do long division? Can we use partial fractions instead?

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It actually says do long division?
Long division is going to get you nowhere in this case. Notice the power of the denominator is bigger than that of the numerator.
Oh that is true
good eye
ok, so how would you do it with partial fractions?
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.
Would you like a step by step?
if you could, that would help a lot!
No problem at all, first: how familiar are you with partial fractions?
i'm not familiar with the name but once i see it i will probably remember..
i don't think we have done partial fractions, i just looked it up on wolfram alpha and it doesn't look familiar
is there any other way to solve it?
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: \[\frac{27x^2}{(3x+4)^3}=\frac{A}{3x+4}+\frac{B}{(3x+4)^2}+\frac{C}{(3x+4)^3} \implies \] \[A(3x+4)^2+B(3x+4)+C=27x^2\]
ok now it looks familiar!!
so how would i get A B and C?
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?
Is it ok if i wrote it:\[27\int\limits_{}^{}1\div(9(3x+4)-8\div9(3x+4)^2+16\div9(3x+4)^3 dx\]
i took out constant first are my numbers correct?
Yeah, it would be just (constant)/27. So I believe so. Do you know how to integrate each of these terms?
well what would i use for my u value if i were to use u-substitution?
so that i only have to substitute once.
For all of them let u=3x+4; (1/3)du=dx
but i will still have to substitute again after
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^^
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.
i got \[(u^2\div2)+(8\div27u)-(8\div27u^2)+ c] is that correct?
\[(u^2/2)+(8/27u)-(8/27u^2)+c\]**
then plug in u-value?

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