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Math Help - Integral Problem

  1. #1
    adnaP
    Guest

    Integral Problem

    I can't figure out how to do this problem.
    <br />
\int \frac {1}{x(x^4+1)}<br />
dx<br />
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    INT.[1 / {x(x^4 +1)}]dx

    Mutiply both numerator and denominator by x^3,

    = INT.[x^3 / {(x^4)(x^4 +1)}] dx
    = INT.[1 / {(x^4)(x^4 +1)}] (x^3 dx) ------------(i)

    Let u = x^4
    So, du = (4x^3)dx

    Then (i) becomes
    = INT.[1 / (u*(u+1))]*(1/4)du
    = (1/4)INT.[1 / u(u+1)]du ------------------(ii)

    By partial fraction:
    1 / u(u+1) = A/u +B/(u+1)
    Multiply both sides by u(u+1),
    1 = A(u+1) +B*u ------------------(a)

    When u = -1, in (a),
    1 = 0 +B(-1)
    So, B = -1

    When u = 0, in (a),
    1 = A(0+1) +0
    So, A = 1

    Hence, (ii) becomes
    = (1/4)INT.[1/u -1/(u+1)]du
    = (1/4)ln(u) -(1/4)ln(u+1) +C ---------(iii)

    Return to u = x^4,
    = (1/4)ln(x^4) -(1/4)ln(x^4 +1) +C
    = ln[(x^4)^(1/4)] -(1/4)ln(x^4 +1) +C
    = ln(x) -(1/4)ln(x^4 +1) +C --------------answer.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
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    Santiago, Chile
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    Quote Originally Posted by adnaP View Post
    <br />
\int \frac {1}{x(x^4+1)}<br />
dx<br />
    Note that

    \frac{1}<br />
{{x\left( {x^4 + 1} \right)}} = \frac{{\left( {x^4 + 1} \right) - x^4 }}<br />
{{x\left( {x^4 + 1} \right)}} = \frac{1}<br />
{x} - \frac{{x^3 }}<br />
{{x^4 + 1}}

    Integrating

    \int {\frac{1}<br />
{{x\left( {x^4 + 1} \right)}}\,dx} = \int {\frac{1}<br />
{x}\,dx} - \frac{1}<br />
{4}\int {\frac{{\left( {x^4 + 1} \right)'}}<br />
{{x^4 + 1}}\,dx} = \ln \left| x \right| - \frac{1}<br />
{4}\ln \left( {x^4 + 1} \right) + k

    From there (if you desire), you may do some of make-up with the natural log.
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