dessyj1
  • dessyj1
Calculus 1. Question in the comments
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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dessyj1
  • dessyj1
\[\int\limits_{none}^{none} \frac{ 11x+4 }{ (3+6x-x ^{2})^{2} } dx\]
dessyj1
  • dessyj1
this is supposed to be an indefinite integral.
dessyj1
  • dessyj1
I cannot easily do u-substitution

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anonymous
  • anonymous
Use 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
  • anonymous
Multiply to eliminate fractions \[11x+4=(Ax+B)(-x^2+6x+3)+Cx+D\]
dessyj1
  • dessyj1
Would you mind showing that to me? I only learned u-substitution.
dessyj1
  • dessyj1
you mean -x^2 instand of x^x right?
anonymous
  • anonymous
yes 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
  • anonymous
Once you have the fractions eliminated, multiply everything out \[11x+4=-Ax^3+6Ax^2+3Ax-Bx^2+6Bx+3B+Cx+D\] Group everything together by the x variable \[11x+4=x^3(-A)+x^2(6A-B)+x(3A+6B+C)+(3B+D)\] *check my multiplication please*
dessyj1
  • dessyj1
checks out.
anonymous
  • anonymous
Now 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 \[0=6A-B\] For x, \[11=3A+6B+C\] For the constant 4 = 3B+D
dessyj1
  • dessyj1
that makes sense
anonymous
  • anonymous
When 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
  • dessyj1
Did you solve the system of equations by going off of the fact that you knew A=0?
anonymous
  • anonymous
yes I did. Ignore this picture for now |dw:1434688610771:dw|
anonymous
  • anonymous
If we use u-substitution, \[u=-x^2+6x+3\] \[du=(-2x+6)dx\] 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
  • anonymous
So now you're taking the integral of|dw:1434689368064:dw|
anonymous
  • anonymous
you can use u-substitution 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
anonymous
  • anonymous
you with me so far?
dessyj1
  • dessyj1
I am.
dessyj1
  • dessyj1
wait.
dessyj1
  • dessyj1
why did you just multiply (-2/-2) to 11?
anonymous
  • anonymous
because the coefficient of x in du is -2 and I needed it to match that.
dessyj1
  • dessyj1
why did it not apply to the 4 as well?
anonymous
  • anonymous
It 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
  • dessyj1
did you try to balance that by adding 37?
dessyj1
  • dessyj1
if that is the case then i am with you.
anonymous
  • anonymous
|dw:1434690426953:dw|
anonymous
  • anonymous
yes, adding 37 balances it
dessyj1
  • dessyj1
forgot the neg 1 exp
dessyj1
  • dessyj1
and how did you know it would balance it?
anonymous
  • anonymous
oh, sharp eye :) |dw:1434690646799:dw|
anonymous
  • anonymous
because 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
  • anonymous
This is what I found in a table for the right integral
1 Attachment
anonymous
  • anonymous
a = -1, b = 6, c = 3, and n = 2
anonymous
  • anonymous
So 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
  • anonymous
|dw:1434691058829:dw|
anonymous
  • anonymous
The top one
1 Attachment
dessyj1
  • dessyj1
is what you just sent a suggestion? can i just use the previous formula?
anonymous
  • anonymous
you 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
  • dessyj1
would the integral at the end of the first be multiplied to its preceding equations?
anonymous
  • anonymous
Just 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
  • dessyj1
so the jumbled mess behind the integral is kind of like a constant?
anonymous
  • anonymous
which part? or which integral?
dessyj1
  • dessyj1
you showed me the first formula
dessyj1
  • dessyj1
and behind the integrand of the integral at the end is a constant.
anonymous
  • anonymous
oh 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
  • anonymous
have you used that second formula yet?
dessyj1
  • dessyj1
yes i have.
dessyj1
  • dessyj1
i ended up with \[\sqrt{48}\] as my discriminant, which is not pretty.
anonymous
  • anonymous
Use the middle formula. I did b²-4ac when their descriptions are for 4ac-b². Looks like you did the same thing I did.
1 Attachment
dessyj1
  • dessyj1
it shows b^2-4ac though...
anonymous
  • anonymous
The discriminant is still the same for the purpose of the integral. |dw:1434692939744:dw|
anonymous
  • anonymous
Above 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
  • dessyj1
oh okay.
dessyj1
  • dessyj1
i used the onw with the ln
dessyj1
  • dessyj1
yeah thats what i got as well.
anonymous
  • anonymous
ok cool
dessyj1
  • dessyj1
+ the squarert
dessyj1
  • dessyj1
|dw:1434693216773:dw|
anonymous
  • anonymous
yes
dessyj1
  • dessyj1
that is not the answer for the right integral is it?
dessyj1
  • dessyj1
it still have to be multiplies by the sum in the first formula right? and by 37
anonymous
  • anonymous
no the answer is the whole thing. |dw:1434693381363:dw|
dessyj1
  • dessyj1
Okay.
dessyj1
  • dessyj1
|dw:1434693703858:dw|
dessyj1
  • dessyj1
|dw:1434693767672:dw|
dessyj1
  • dessyj1
I have a calc 1 final in about 13 hours and i need to sleep.
dessyj1
  • dessyj1
Thank you for all your help. I really appreciate it.
anonymous
  • anonymous
you're welcome. good luck! You should be fine. Any prof who puts something this insane on a final is a sadist.
anonymous
  • anonymous
\[\frac{ 37(-2x+6) }{ -48(-x^2+6x+3) }+\frac{ 37 }{ 24\sqrt{48} }\ln|\frac{ -2x+6+\sqrt{48} }{ -2x+6-\sqrt{48} }|+\frac{ 11 }{ 2\sqrt{-x^2+6x+3}}+C\]

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