Calculus 1. Question in the comments

- dessyj1

Calculus 1. Question in the comments

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- dessyj1

\[\int\limits_{none}^{none} \frac{ 11x+4 }{ (3+6x-x ^{2})^{2} } dx\]

- dessyj1

this is supposed to be an indefinite integral.

- dessyj1

I cannot easily do u-substitution

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## More answers

- 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

Multiply to eliminate fractions
\[11x+4=(Ax+B)(-x^2+6x+3)+Cx+D\]

- dessyj1

Would you mind showing that to me? I only learned u-substitution.

- dessyj1

you mean -x^2 instand of x^x right?

- 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

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

checks out.

- 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

that makes sense

- 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

Did you solve the system of equations by going off of the fact that you knew A=0?

- anonymous

yes I did. Ignore this picture for now
|dw:1434688610771:dw|

- 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

So now you're taking the integral of|dw:1434689368064:dw|

- 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

you with me so far?

- dessyj1

I am.

- dessyj1

wait.

- dessyj1

why did you just multiply (-2/-2) to 11?

- anonymous

because the coefficient of x in du is -2 and I needed it to match that.

- dessyj1

why did it not apply to the 4 as well?

- 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

did you try to balance that by adding 37?

- dessyj1

if that is the case then i am with you.

- anonymous

|dw:1434690426953:dw|

- anonymous

yes, adding 37 balances it

- dessyj1

forgot the neg 1 exp

- dessyj1

and how did you know it would balance it?

- anonymous

oh, sharp eye :)
|dw:1434690646799:dw|

- 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

This is what I found in a table for the right integral

##### 1 Attachment

- anonymous

a = -1, b = 6, c = 3, and n = 2

- 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

|dw:1434691058829:dw|

- anonymous

The top one

##### 1 Attachment

- dessyj1

is what you just sent a suggestion? can i just use the previous formula?

- 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

would the integral at the end of the first be multiplied to its preceding equations?

- 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

so the jumbled mess behind the integral is kind of like a constant?

- anonymous

which part? or which integral?

- dessyj1

you showed me the first formula

- dessyj1

and behind the integrand of the integral at the end is a constant.

- 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

have you used that second formula yet?

- dessyj1

yes i have.

- dessyj1

i ended up with \[\sqrt{48}\] as my discriminant, which is not pretty.

- 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

it shows b^2-4ac though...

- anonymous

The discriminant is still the same for the purpose of the integral.
|dw:1434692939744:dw|

- 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

oh okay.

- dessyj1

i used the onw with the ln

- dessyj1

yeah thats what i got as well.

- anonymous

ok cool

- dessyj1

+ the squarert

- dessyj1

|dw:1434693216773:dw|

- anonymous

yes

- dessyj1

that is not the answer for the right integral is it?

- dessyj1

it still have to be multiplies by the sum in the first formula right? and by 37

- anonymous

no the answer is the whole thing.
|dw:1434693381363:dw|

- dessyj1

Okay.

- dessyj1

|dw:1434693703858:dw|

- dessyj1

|dw:1434693767672:dw|

- dessyj1

I have a calc 1 final in about 13 hours and i need to sleep.

- dessyj1

Thank you for all your help. I really appreciate it.

- anonymous

you're welcome. good luck! You should be fine. Any prof who puts something this insane on a final is a sadist.

- 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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