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Math Help - 3 Questions

  1. #1
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    3 Questions

    Use partial fractions to find:
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx
    \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx
    =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid
    =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid
    Answer is supposed to be 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid

    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx
    \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}
    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx
    =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid
    =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid
    Answer is supposed to be \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid

    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du
    \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}
    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du
    =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid
    Answer is supposed to be 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid
    I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    Use partial fractions to find:
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx
    \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx
    =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid
    =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid
    Answer is supposed to be 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid

    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx
    \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}
    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx
    =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid
    =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid
    Answer is supposed to be \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid

    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du
    \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}
    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du
    =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid
    Answer is supposed to be 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid
    I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
    Thanks

    In the first one, everything is right up to this step...

    \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx

    I don't know where you got the 13\ln{|3-x|} part...


    It should be

     = -12\ln{|2 - x|} + 36\ln{|3 - x|} - 24\ln{|4 - x|} + C

     = 12\left[-\ln{|2 - x|} + 3\ln{|3 - x|} - 2\ln{|4 - x|}\right] + C

     = 12\left[-\ln{|2 - x|} + \ln{|3 - x|^3} - \ln{|4 - x|^2}\right] + C

     = 12\left[\ln{\frac{|3 - x|^3}{|2 - x||4 - x|^2}}\right] + C

     = 12\ln{\left|\frac{(3 - x)^3}{(2 - x)(4 - x)^2}\right|} +C.
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  3. #3
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    Quote Originally Posted by arze View Post
    Use partial fractions to find:
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx
    \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx
    =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid
    =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid
    Answer is supposed to be 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid

    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx
    \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}
    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx
    =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid
    =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid
    Answer is supposed to be \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid

    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du
    \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}
    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du
    =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid
    Answer is supposed to be 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid
    I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
    Thanks
    For Q.2, Again, correct up to

    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx

    This should be

     = -\frac{7}{2}\ln{|2x - 1|} + \frac{7}{2}\ln{|x - 1|} + \frac{1}{2}\ln{|x + 1|} + C

     = \frac{1}{2}\left[-7\ln{|2x - 1|} + 7\ln{|x - 1|} + \ln{|x + 1|}\right] + C

     = \frac{1}{2}\left[-\ln{|2x - 1|^7} + \ln{|x - 1|^7} + \ln{|x + 1|}\right] + C

     = \frac{1}{2}\left[\ln{\left(\frac{|x - 1|^7|x + 1|}{|2x - 1|^7}\right)}\right]+C

     = \frac{1}{2}\ln{\left|\frac{(x -1)^7(x + 1)}{(2x - 1)^7}\right|} + C.
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  4. #4
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    for your first one at the very least, your fractions are wrong.
    I'm pretty sure it's supposed to be 12/(2-x)+-36/(3-x)+24/(4-x)

    going to take a bit to check the others.
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  5. #5
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    Quote Originally Posted by arze View Post
    Use partial fractions to find:
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx
    \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}
    \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx
    =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid
    =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid
    Answer is supposed to be 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid

    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx
    \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}
    \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx
    =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid
    =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid
    Answer is supposed to be \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid

    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du
    \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}
    \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du
    =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid
    Answer is supposed to be 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid
    I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
    Thanks
    For Q.3, you have done everything right except the final integration.

    You made the same mistake in ALL of the questions.


    You need to remember that

    \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\ln{|ax + b|} + C.


    You keep forgetting to take out the common factor of \frac{1}{a}...
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  6. #6
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    Quote Originally Posted by Prove It View Post
    In the first one, everything is right up to this step...

    \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx

    I don't know where you got the 13\ln{|3-x|} part...


    It should be

     = -12\ln{|2 - x|} + 36\ln{|3 - x|} - 24\ln{|4 - x|} + C

     = 12\left[-\ln{|2 - x|} + 3\ln{|3 - x|} - 2\ln{|4 - x|}\right] + C

     = 12\left[-\ln{|2 - x|} + \ln{|3 - x|^3} - \ln{|4 - x|^2}\right] + C

     = 12\left[\ln{\frac{|3 - x|^3}{|2 - x||4 - x|^2}}\right] + C

     = 12\ln{\left|\frac{(3 - x)^3}{(2 - x)(4 - x)^2}\right|} +C.
    The 13 was a typo. Thanks!
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  7. #7
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    Now I see where I got it wrong. Thanks much
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