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

1. anonymous

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

2. anonymous

oh ok

3. myininaya

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

4. myininaya

It actually says do long division?

5. 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.

6. myininaya

Oh that is true

7. myininaya

good eye

8. anonymous

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

9. 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.

10. anonymous

Would you like a step by step?

11. anonymous

if you could, that would help a lot!

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

is there any other way to solve it?

16. 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: $\frac{27x^2}{(3x+4)^3}=\frac{A}{3x+4}+\frac{B}{(3x+4)^2}+\frac{C}{(3x+4)^3} \implies$ $A(3x+4)^2+B(3x+4)+C=27x^2$

17. anonymous

ok now it looks familiar!!

18. anonymous

so how would i get A B and C?

19. 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?

20. anonymous

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

21. anonymous

i took out constant first are my numbers correct?

22. anonymous

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

23. anonymous

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

24. anonymous

so that i only have to substitute once.

25. anonymous

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

26. anonymous

but i will still have to substitute again after

27. 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^^

28. 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.

29. anonymous

i got $(u^2\div2)+(8\div27u)-(8\div27u^2)+ c] is that correct? 30. anonymous \[(u^2/2)+(8/27u)-(8/27u^2)+c$**

31. anonymous

then plug in u-value?