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Math Help - Find Indefinite Integral: Complete the Square

  1. #1
    Super Member nycmath's Avatar
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    Find Indefinite Integral: Complete the Square

    Must I complete the square of the denominator? If so, after completing the square my answer is (x+1)^2 + (1/2).

    I then set a to be sqrt(1/2) and u to be x+1, which means dx = du. Is this right so far.

    Can someone rotate this picture for easy reading?
    Attached Thumbnails Attached Thumbnails Find Indefinite Integral: Complete the Square-cam00257.jpg  
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    Re: Find Indefinite Integral: Complete the Square

    Quote Originally Posted by nycmath View Post
    Must I complete the square of the denominator? If so, after completing the square my answer is (x+1)^2 + (1/2).

    I then set a to be sqrt(1/2) and u to be x+1, which means dx = du. Is this right so far.

    Can someone rotate this picture for easy reading?
    Find Indefinite Integral: Complete the Square-clipboard01.jpg
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    Re: Find Indefinite Integral: Complete the Square

    Quote Originally Posted by romsek View Post
    Click image for larger version. 

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    $\dfrac{1}{1-4x-2x^2}=$

    $\dfrac{1}{2}\dfrac{1}{1/2 - 2x - x^2}=$

    $\dfrac{1}{2}\dfrac{1}{1/2 - (x+1)^2+1}=$

    $\dfrac{1}{2}\dfrac{1}{3/2-(x+1)^2}=$

    $\dfrac{1}{2}\dfrac{1}{(\sqrt{3/2})^2-(x+1)^2}$
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    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    Where did the 1/2 in front of the fraction come from?
    Is 1-4x-2x^2 the same when written
    -2x^2-4x+1? This is what I assumed the original denominator to be by simply rearranging the terms before completing the square. Are you saying that I completed the square wrongly? Good night. I will check your reply in the morning.
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    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    romsek,

    I let a = sqrt(3/2) which can be written as sqrt(6)/2. I let u be x+1, which means
    dx = du. When I plug those values into the formula in the picture, my answer is the following:

    [1/sqrt(6) * ln |[sqrt(6)/2] + x+1]/sqrt(6)/2 - (x-1)] + C.

    What am I doing wrong?
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    Re: Find Indefinite Integral: Complete the Square

    Quote Originally Posted by nycmath View Post
    romsek,

    I let a = sqrt(3/2) which can be written as sqrt(6)/2. I let u be x+1, which means
    dx = du. When I plug those values into the formula in the picture, my answer is the following:

    [1/sqrt(6) * ln |[sqrt(6)/2] + x+1]/sqrt(6)/2 - (x-1)] + C.

    What am I doing wrong?
    you left out the factor of 1/2 out front for one.

    I just worked it all out and matched wolfram's answer using $a=\sqrt{\frac{3}{2}}$ so it does work.

    Rework your algebra.
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    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    I see now that my confusion stems from not knowing where the factor of 1/2 came from.
    Where does the factor of 1/2 come from?
    Can you break this down for me? I will work on the algebra part.
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    Re: Find Indefinite Integral: Complete the Square

    Quote Originally Posted by nycmath View Post
    I see now that my confusion stems from not knowing where the factor of 1/2 came from.
    Where does the factor of 1/2 come from?
    Can you break this down for me? I will work on the algebra part.
    $\dfrac{1}{1-4x-2x^2}=\dfrac{1}{2(1/2-2x-x^2)}=\dfrac{1}{2}\dfrac{1}{1/2-2x-x^2}$
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    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    I see what happened. I divided across by -2 when I should have factored out a 2 instead. By factoring out 2, the numerator being 1, there is my factor 1/2 in front. Great work, romsek.

    Tomorrow I begin learning how to find the area between two curves. Try to understand that I am learning calculus material outside a classroom setting. I hope you do not think I am nuts but math has been my passion for many years.

    I am 48 years old. It makes zero sense to take out a school loan for the purpose of majoring in math at this late stage in my life. But, we are never too old to do a self-study of any topic, right? What is your input in terms of learning just to know how to do something?
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    Re: Find Indefinite Integral: Complete the Square

    I keeps your brain healthy and that's reason enough even if you don't end up being another Grigory Perelman.
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    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    romsek,

    Thanks. Yes, math helps me to continue paying attention to detail. My job has nothing to do with math. I have two CUNY degrees in areas other than math. I took precalculus in 1993 and got an A minus. I should have continued in math but made a wrong a choice. I also have an online degree in Religious Education with concentration on eschatology. My friends and family see no purpose in studying calculus because they say it is not practical. They say I will never be James Stewart. I am not trying to be James Stewart. I just love math as crazy as that sounds. I am divorced. I am not popular with the ladies. So, why not learn math, right?
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    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    romsek,

    I did the algebra and still got the wrong answer. My answer:

    1/(2*sqrt{6}) ln |sqrt(6)/2 + (x+1) divided by sqrt(6)/2 - (x+1)| + C

    I give up here. Can you finish this question in steps? It will help me as a guide for future similar questions.
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  13. #13
    Super Member nycmath's Avatar
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    Re: Find Indefinite Integral: Complete the Square

    According to the Wolfram Math website, my answer is correct.
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