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Math Help - Difficult Integral

  1. #1
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    Difficult Integral

    Hi all,

    Im studying calculus now and Ive come across this question in a workbook and im struggling with it. Its to find this integral:

    ∫ 7x / x + 1 dx

    If someone could show me it would be a great help and much of a relief.

    Thanks in advance,

    Mitch
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  2. #2
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    Quote Originally Posted by Mitch281008 View Post
    Hi all,

    Im studying calculus now and Ive come across this question in a workbook and im struggling with it. Its to find this integral:

    ∫ 7x / x + 1 dx

    If someone could show me it would be a great help and much of a relief.

    Thanks in advance,

    Mitch
    Make the substitition u=x^3+1 \implies du=3x^2dx

    This gives

    \int \frac{7x^2}{x^3+1}dx=\frac{7}{3}\int\frac{1}{u}du

    Now just integrate from here
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  3. #3
    Newbie Riyzar's Avatar
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    Have you used U substitution?

    Try setting u=x^3

    Then:  du=3x^2dx

    -> \frac{7du}{3} = 7x^2

    The integral then becomes: \frac{7}{3}\int{\frac{1}{u}}du +\int1dx
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Riyzar View Post
    Have you used U substitution?

    Try setting u=x^3

    Then:  du=3x^2dx

    -> \frac{7du}{3} = 7x^2

    The integral then becomes: \frac{7}{3}\int{\frac{1}{u}}du +\int1dx
    You are supposing u = x^3

    NOTE THAT you have to suppose

    u = {x^3}+ 1
    then, du = 3x^2 dx

    your integral is :

    \frac{7}{3}\int{\frac{1}{u}}du

    = \frac{7}{3} log(u) + C

    substitute u = {x^3}+ 1

    and you're are smart enough to write the answer
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  5. #5
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    Ok thanks guys.

    It was one of those questions where i could not get an answer no matter what i tried.

    i appreciate the help

    cheers,
    Mitch
    Last edited by Mitch281008; March 13th 2010 at 08:46 PM.
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  6. #6
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    Ok I have another problem, this one ive solved but i just want to check the answer to see if ive got it.

    ∫ 5e^(3x+1) = 5 (e^(3x+1)) / (3x+1) + c

    i hope thats clear, its annoying not being able to make fractions look like fractions lol.

    thanks for the support,

    Mitch
    Last edited by Mitch281008; March 14th 2010 at 01:17 AM.
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Mitch281008 View Post
    Ok I have another problem, this one ive solved but i just want to check the answer to see if ive got it.

    ∫ 5e^(3x+1) = 5 (e^(3x+1)) / (3x+1) + c

    i hope thats clear, its annoying not being able to make fractions look like fractions lol.

    thanks for the support,

    Mitch
    mitch, your answer looks erroneus to me

    \int 5e^{3x+1} dx

    let u = 3x+1

    then du = 3dx

    so, dx = \frac{du}{3}

    you have,

    \int \frac{5e^{u}du}{3}


    = \frac{5e^u}{3}+C


    substitute  u = 3x+1

    you get,

    \frac{5e^{3x+1}}{3}+C
    Last edited by harish21; March 14th 2010 at 03:16 AM.
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