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Thread: Integral. Verification of decision

  1. #1
    Senior Member DeMath's Avatar
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    Integral. Verification of decision

    Please check my decision.
    Can I find this integral by more simple method?
    I think that I used bad replacement of the variable of integration.

    $\displaystyle
    \int {\frac{{dx}}
    {{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} = \left\{ \begin{gathered}
    x = t^6 , \hfill \\
    dx = 6t^5 dt \hfill \\
    \end{gathered} \right\} = 6\int {\frac{{dt}}
    {{t\left( {2t^3 + t^2 + 1} \right)}}.}
    $

    $\displaystyle
    \begin{gathered}
    2t^3 + t^2 + 1 = 2t^3 + 2 + t^2 - 1 = 2\left( {t^3 + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \hfill \\
    = 2\left( {t + 1} \right)\left( {t^2 - t + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \left( {t + 1} \right)\left( {2t^2 - t + 1} \right). \hfill \\
    \end{gathered}
    $

    $\displaystyle \begin{gathered}
    \frac{1}
    {{t\left( {t + 1} \right)\left( {2t^2 - t + 1} \right)}} = \frac{A}
    {t} + \frac{B}
    {{t + 1}} + \frac{{Ct + D}}
    {{2t^2 - t + 1}} \Leftrightarrow \hfill \\
    \Leftrightarrow 1 = A\left( {t + 1} \right)\left( {2t^2 - t + 1} \right) + B\left( {2t^2 - t + 1} \right)t + \left( {Ct + D} \right)\left( {t + 1} \right)t = \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    = 2At^3 + At^2 + A + 2Bt^3 - Bt^2 + Bt + Ct^3 + Ct^2 + Dt^2 + Dt = \hfill \\
    = \left( {2A + 2B + C} \right)t^3 + \left( {A - B + C + D} \right)t^2 + \left( {B + D} \right)t + A{\text{ }} \Rightarrow \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \Rightarrow \left\{ \begin{gathered}
    A = 1, \hfill \\
    B + D = 0, \hfill \\
    2A + 2B + C = 0, \hfill \\
    A - B + C + D = 0; \hfill \\
    \end{gathered} \right.
    $

    $\displaystyle
    \Leftrightarrow \left\{ \begin{gathered}
    A = 1, \hfill \\
    B = - D, \hfill \\
    2D - C = 2, \hfill \\
    2D + C = - 1; \hfill \\
    \end{gathered} \right.
    $

    $\displaystyle
    \Leftrightarrow \left\{ \begin{gathered}
    A = 1, \hfill \\
    B = - {1 \mathord{\left/
    {\vphantom {1 {4,}}} \right.
    \kern-\nulldelimiterspace} {4,}} \hfill \\
    D = {1 \mathord{\left/
    {\vphantom {1 4}} \right.
    \kern-\nulldelimiterspace} 4}, \hfill \\
    C = - {3 \mathord{\left/
    {\vphantom {3 {2.}}} \right.
    \kern-\nulldelimiterspace} {2.}} \hfill \\
    \end{gathered} \right.
    $

    $\displaystyle
    \begin{gathered}
    6\int {\frac{{dt}}
    {{t\left( {2t^3 + t^2 + 1} \right)}}} = 6\int {\frac{1}
    {t}dt} - \frac{3}
    {2}\int {\frac{1}
    {{t + 1}}} dt - \frac{3}
    {2}\int {\frac{{6t - 1}}
    {{2t^2 - t + 1}}} dt = I_1 - I_2 - I_3 . \hfill \\
    I_1 = 6\int {\frac{1}
    {t}dt} = 6\ln t + C = 6\ln \left( {\sqrt[6]{x}} \right) + C = \ln x + C. \hfill \\
    \end{gathered}
    $

    $\displaystyle
    I_2 = \frac{3}
    {2}\int {\frac{1}
    {{t + 1}}} dt = \frac{3}
    {2}\ln \left( {t + 1} \right) + C = \frac{3}
    {2}\ln \left( {\sqrt[6]{x} + 1} \right) + C.
    $

    $\displaystyle
    I_3 = \frac{3}
    {2}\int {\frac{{6t - 1}}
    {{2t^2 - t + 1}}} dt = \frac{9}
    {4}\int {\frac{{4t - {2 \mathord{\left/
    {\vphantom {2 3}} \right.
    \kern-\nulldelimiterspace} 3}}}
    {{2t^2 - t + 1}}dt} = \frac{9}
    {4}\int {\frac{{4t - 1 + 1 - {2 \mathord{\left/
    {\vphantom {2 3}} \right.
    \kern-\nulldelimiterspace} 3}}}
    {{2t^2 - t + 1}}dt} = $


    $\displaystyle
    \begin{gathered}
    = \frac{9}
    {4}\int {\frac{{4t - 1}}
    {{2t^2 - t + 1}}} dt + \frac{3}
    {4}\int {\frac{{dt}}
    {{2t^2 - t + 1}}} . \hfill \\
    \frac{9}
    {4}\int {\frac{{4t - 1}}
    {{2t^2 - t + 1}}dt} = \frac{9}
    {4}\int {\frac{{d\left( {2t^2 - t + 1} \right)}}
    {{2t^2 - t + 1}}} = \frac{9}
    {4}\ln \left( {2t^2 - t + 1} \right) + C. \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \frac{3}
    {4}\int {\frac{{dt}}
    {{2t^2 - t + 1}}} = 6\int {\frac{{dt}}
    {{16t^2 - 8t + 1 + 7}}} = 6\int {\frac{{dt}}
    {{\left( {4t - 1} \right)^2 + 7}}} =
    $

    $\displaystyle
    = \frac{6}
    {7}\int {\frac{{dt}}
    {{\left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
    {{2\sqrt 7 }}\int {\frac{{d\left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right)}}
    {{\left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right) + C.
    $

    $\displaystyle
    \begin{gathered}
    I_3 = \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right) + \frac{9}
    {4}\ln \left( {2t^2 - t + 1} \right) + C = \hfill \\
    = \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
    {{\sqrt 7 }}} \right) + \frac{9}
    {4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) + C. \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    \boxed{\int {\frac{{dx}}
    {{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} } = \ln x - \frac{3}
    {2}\ln \left( {\sqrt[6]{x} + 1} \right) - \frac{9}
    {4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) - \hfill \\
    - \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
    {{\sqrt 7 }}} \right) + C. \hfill \\
    \end{gathered}
    $
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  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by DeMath View Post
    Please check my decision.
    Can I find this integral by more simple method?
    I think that I used bad replacement of the variable of integration.

    $\displaystyle
    \int {\frac{{dx}}
    {{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} = \left\{ \begin{gathered}
    x = t^6 , \hfill \\
    dx = 6t^5 dt \hfill \\
    \end{gathered} \right\} = 6\int {\frac{{dt}}
    {{t\left( {2t^3 + t^2 + 1} \right)}}.}
    $

    $\displaystyle
    \begin{gathered}
    2t^3 + t^2 + 1 = 2t^3 + 2 + t^2 - 1 = 2\left( {t^3 + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \hfill \\
    = 2\left( {t + 1} \right)\left( {t^2 - t + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \left( {t + 1} \right)\left( {2t^2 - t + 1} \right). \hfill \\
    \end{gathered}
    $

    $\displaystyle \begin{gathered}
    \frac{1}
    {{t\left( {t + 1} \right)\left( {2t^2 - t + 1} \right)}} = \frac{A}
    {t} + \frac{B}
    {{t + 1}} + \frac{{Ct + D}}
    {{2t^2 - t + 1}} \Leftrightarrow \hfill \\
    \Leftrightarrow 1 = A\left( {t + 1} \right)\left( {2t^2 - t + 1} \right) + B\left( {2t^2 - t + 1} \right)t + \left( {Ct + D} \right)\left( {t + 1} \right)t= \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    = 2At^3 + At^2 + A + 2Bt^3 - Bt^2 + Bt + Ct^3 + Ct^2 + Dt^2 + Dt = \hfill \\
    = \left( {2A + 2B + C} \right)t^3 + \left( {A - B + C + D} \right)t^2 + \left( {B + D} \right)t + A{\text{ }} \Rightarrow \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \Rightarrow \left\{ \begin{gathered}
    A = 1, \hfill \\
    B + D = 0, \hfill \\
    2A + 2B + C = 0, \hfill \\
    A - B + C + D = 0; \hfill \\
    \end{gathered} \right.
    $

    $\displaystyle
    \Leftrightarrow \left\{ \begin{gathered}
    A = 1, \hfill \\
    B = - D, \hfill \\
    2D - C = 2, \hfill \\
    2D + C = - 1; \hfill \\
    \end{gathered} \right.
    $

    $\displaystyle
    \Leftrightarrow \left\{ \begin{gathered}
    A = 1, \hfill \\
    B = - {1 \mathord{\left/
    {\vphantom {1 {4,}}} \right.
    \kern-\nulldelimiterspace} {4,}} \hfill \\
    D = {1 \mathord{\left/
    {\vphantom {1 4}} \right.
    \kern-\nulldelimiterspace} 4}, \hfill \\
    C = - {3 \mathord{\left/
    {\vphantom {3 {2.}}} \right.
    \kern-\nulldelimiterspace} {2.}} \hfill \\
    \end{gathered} \right.
    $

    $\displaystyle
    \begin{gathered}
    6\int {\frac{{dt}}
    {{t\left( {2t^3 + t^2 + 1} \right)}}} = 6\int {\frac{1}
    {t}dt} - \frac{3}
    {2}\int {\frac{1}
    {{t + 1}}} dt - \frac{3}
    {2}\int {\frac{{6t - 1}}
    {{2t^2 - t + 1}}} dt = I_1 - I_2 - I_3 . \hfill \\
    I_1 = 6\int {\frac{1}
    {t}dt} = 6\ln t + C = 6\ln \left( {\sqrt[6]{x}} \right) + C = \ln x + C. \hfill \\
    \end{gathered}
    $

    $\displaystyle
    I_2 = \frac{3}
    {2}\int {\frac{1}
    {{t + 1}}} dt = \frac{3}
    {2}\ln \left( {t + 1} \right) + C = \frac{3}
    {2}\ln \left( {\sqrt[6]{x} + 1} \right) + C.
    $

    $\displaystyle
    I_3 = \frac{3}
    {2}\int {\frac{{6t - 1}}
    {{2t^2 - t + 1}}} dt = \frac{9}
    {4}\int {\frac{{4t - {2 \mathord{\left/
    {\vphantom {2 3}} \right.
    \kern-\nulldelimiterspace} 3}}}
    {{2t^2 - t + 1}}dt} = \frac{9}
    {4}\int {\frac{{4t - 1 + 1 - {2 \mathord{\left/
    {\vphantom {2 3}} \right.
    \kern-\nulldelimiterspace} 3}}}
    {{2t^2 - t + 1}}dt} = $


    $\displaystyle
    \begin{gathered}
    = \frac{9}
    {4}\int {\frac{{4t - 1}}
    {{2t^2 - t + 1}}} dt + \frac{3}
    {4}\int {\frac{{dt}}
    {{2t^2 - t + 1}}} . \hfill \\
    \frac{9}
    {4}\int {\frac{{4t - 1}}
    {{2t^2 - t + 1}}dt} = \frac{9}
    {4}\int {\frac{{d\left( {2t^2 - t + 1} \right)}}
    {{2t^2 - t + 1}}} = \frac{9}
    {4}\ln \left( {2t^2 - t + 1} \right) + C. \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \frac{3}
    {4}\int {\frac{{dt}}
    {{2t^2 - t + 1}}} = 6\int {\frac{{dt}}
    {{16t^2 - 8t + 1 + 7}}} = 6\int {\frac{{dt}}
    {{\left( {4t - 1} \right)^2 + 7}}} =
    $

    $\displaystyle
    = \frac{6}
    {7}\int {\frac{{dt}}
    {{\left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
    {{2\sqrt 7 }}\int {\frac{{d\left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right)}}
    {{\left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right) + C.
    $

    $\displaystyle
    \begin{gathered}
    I_3 = \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
    {{\sqrt 7 }}} \right) + \frac{9}
    {4}\ln \left( {2t^2 - t + 1} \right) + C = \hfill \\
    = \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
    {{\sqrt 7 }}} \right) + \frac{9}
    {4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) + C. \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    \boxed{\int {\frac{{dx}}
    {{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} } = \ln x - \frac{3}
    {2}\ln \left( {\sqrt[6]{x} + 1} \right) - \frac{9}
    {4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) - \hfill \\
    - \frac{3}
    {{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
    {{\sqrt 7 }}} \right) + C. \hfill \\
    \end{gathered}
    $
    there's nothing wrong with your solution. good work!
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