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dessyj1
 one year ago
Calculus 1. Question in the comments
dessyj1
 one year ago
Calculus 1. Question in the comments

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dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{none}^{none} \frac{ 11x+4 }{ (3+6xx ^{2})^{2} } dx\]

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0this is supposed to be an indefinite integral.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0I cannot easily do usubstitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use partial fraction decomposition \[\frac{ 11x+4 }{(x^2+6x+3)^2 }=\frac{ Ax+B }{ x^x+6x+3 }+\frac{ Cx+D }{ (x^2+6x+3)^2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Multiply to eliminate fractions \[11x+4=(Ax+B)(x^2+6x+3)+Cx+D\]

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0Would you mind showing that to me? I only learned usubstitution.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0you mean x^2 instand of x^x right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes I do mean x². Read up on it here: http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx There's a table in the middle of the page that shows what format the decomposed fraction is supposed to do. I'll work through this one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once you have the fractions eliminated, multiply everything out \[11x+4=Ax^3+6Ax^2+3AxBx^2+6Bx+3B+Cx+D\] Group everything together by the x variable \[11x+4=x^3(A)+x^2(6AB)+x(3A+6B+C)+(3B+D)\] *check my multiplication please*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now you can equate coefficients to create a system of equations. On the left, there is no x³ term and on the right the coefficient of x³ is A, so the equation is \[0=A\] For x², it's \[0=6AB\] For x, \[11=3A+6B+C\] For the constant 4 = 3B+D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When I solve the system I get A = 0, B = 0, C = 11, and D = 4, which puts us right where we started :/. PFD is usually a solid method for integrating rational functions. Let me see if I can figure another method

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0Did you solve the system of equations by going off of the fact that you knew A=0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes I did. Ignore this picture for now dw:1434688610771:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we use usubstitution, \[u=x^2+6x+3\] \[du=(2x+6)dx\] We need to make the numerator 11x+4 look like 2x+6. Start by multiplying by 1 in the form of 2/2 \[11(\frac{ 2 }{ 2}) x+4\]\[\frac{ 11 }{ 2 }(2x)+4\] I'm going to add 6 inside the parentheses, which is the same as adding 33 to the overall expression. I'll have to add 33 outside so the expression's value doesn't change \[\frac{ 11 }{ 2 }(2x+6)+37\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So now you're taking the integral ofdw:1434689368064:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can use usubstitution for the integral on the left. And I'm going to check if I can find the one on the right in an integral table

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0why did you just multiply (2/2) to 11?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because the coefficient of x in du is 2 and I needed it to match that.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0why did it not apply to the 4 as well?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It wouldn't have had any effect. The 2's would have just cancelled. Technically, they cancel when multiplied by 11, but I didn't because I want the coefficient to be 2.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0did you try to balance that by adding 37?

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0if that is the case then i am with you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434690426953:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, adding 37 balances it

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0and how did you know it would balance it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, sharp eye :) dw:1434690646799:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because I added 6 inside the parentheses. Distributing 11/2 means I added 33 to the whole thing. So I had to add 33 to keep the value the same. Then 4 + 33 = 37

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is what I found in a table for the right integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a = 1, b = 6, c = 3, and n = 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So once you fill in all the constants and do what looks to be an obnoxious amount of multiplication, all that's left is to solve the integral on the right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434691058829:dw

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0is what you just sent a suggestion? can i just use the previous formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need them both. If you notice, the first one has and integral toward the end of it. The second one is just the solution to that integral

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0would the integral at the end of the first be multiplied to its preceding equations?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just the part between the plus and integral sign. I personally wouldn't attempt to simplify it. Put the solution to the integral in brackets and write that part outside

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0so the jumbled mess behind the integral is kind of like a constant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which part? or which integral?

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0you showed me the first formula

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0and behind the integrand of the integral at the end is a constant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. it's not a constant because the x's are still there. The coefficient of the integral works out to being a constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0have you used that second formula yet?

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0i ended up with \[\sqrt{48}\] as my discriminant, which is not pretty.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use the middle formula. I did b²4ac when their descriptions are for 4acb². Looks like you did the same thing I did.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0it shows b^24ac though...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The discriminant is still the same for the purpose of the integral. dw:1434692939744:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Above I told you to use the formula with arctan, when we should use the one with ln. Just wanted to make sure we're using the same one

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0i used the onw with the ln

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0yeah thats what i got as well.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0that is not the answer for the right integral is it?

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0it still have to be multiplies by the sum in the first formula right? and by 37

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no the answer is the whole thing. dw:1434693381363:dw

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0I have a calc 1 final in about 13 hours and i need to sleep.

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for all your help. I really appreciate it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're welcome. good luck! You should be fine. Any prof who puts something this insane on a final is a sadist.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 37(2x+6) }{ 48(x^2+6x+3) }+\frac{ 37 }{ 24\sqrt{48} }\ln\frac{ 2x+6+\sqrt{48} }{ 2x+6\sqrt{48} }+\frac{ 11 }{ 2\sqrt{x^2+6x+3}}+C\]