Question

find the integral using long division...?

Answer 1

\[\int\limits_{}^{}(27x^2)\div (3x+4)^3 dx\]

Answer 2

oh ok

Answer 3

Do we have to do long division? Can we use partial fractions instead?

Answer 4

It actually says do long division?

Answer 5

Long division is going to get you nowhere in this case. Notice the power of the denominator is bigger than that of the numerator.

Answer 6

Oh that is true

Answer 7

good eye

Answer 8

ok, so how would you do it with partial fractions?

Answer 9

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.

Answer 10

Would you like a step by step?

Answer 11

if you could, that would help a lot!

Answer 12

No problem at all, first: how familiar are you with partial fractions?

Answer 13

i'm not familiar with the name but once i see it i will probably remember..

Answer 14

i don't think we have done partial fractions, i just looked it up on wolfram alpha and it doesn't look familiar

Answer 15

is there any other way to solve it?

Answer 16

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:
\[\frac{27x^2}{(3x+4)^3}=\frac{A}{3x+4}+\frac{B}{(3x+4)^2}+\frac{C}{(3x+4)^3} \implies \]
\[A(3x+4)^2+B(3x+4)+C=27x^2\]

Answer 17

ok now it looks familiar!!

Answer 18

so how would i get A B and C?

Answer 19

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?

Answer 20

Is it ok if i wrote it:\[27\int\limits_{}^{}1\div(9(3x+4)-8\div9(3x+4)^2+16\div9(3x+4)^3 dx\]

Answer 21

i took out constant first
are my numbers correct?

Answer 22

Yeah, it would be just (constant)/27. So I believe so. Do you know how to integrate each of these terms?

Answer 23

well what would i use for my u value if i were to use u-substitution?

Answer 24

so that i only have to substitute once.

Answer 25

For all of them let u=3x+4; (1/3)du=dx

Answer 26

but i will still have to substitute again after

Answer 27

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^^

Answer 28

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.

Answer 29

i got \[(u^2\div2)+(8\div27u)-(8\div27u^2)+ c] is that correct?

Answer 30

\[(u^2/2)+(8/27u)-(8/27u^2)+c\]**

Answer 31

then plug in u-value?