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Math Help - 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.

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

    <br />
\begin{gathered}<br />
  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 \\<br />
   = 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 \\ <br />
\end{gathered} <br />

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

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

    <br />
 \Rightarrow \left\{ \begin{gathered}<br />
  A = 1, \hfill \\<br />
  B + D = 0, \hfill \\<br />
  2A + 2B + C = 0, \hfill \\<br />
  A - B + C + D = 0; \hfill \\ <br />
\end{gathered}  \right.<br />

    <br />
 \Leftrightarrow \left\{ \begin{gathered}<br />
  A = 1, \hfill \\<br />
  B =  - D, \hfill \\<br />
  2D - C = 2, \hfill \\<br />
  2D + C =  - 1; \hfill \\ <br />
\end{gathered}  \right.<br />

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

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

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

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


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

    <br />
\frac{3}<br />
{4}\int {\frac{{dt}}<br />
{{2t^2  - t + 1}}}  = 6\int {\frac{{dt}}<br />
{{16t^2  - 8t + 1 + 7}}}  = 6\int {\frac{{dt}}<br />
{{\left( {4t - 1} \right)^2  + 7}}}  = <br />

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

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

    <br />
\begin{gathered}<br />
  \boxed{\int {\frac{{dx}}<br />
{{x\left( {1 + 2\sqrt x  + \sqrt[3]{x}} \right)}}} } = \ln x - \frac{3}<br />
{2}\ln \left( {\sqrt[6]{x} + 1} \right) - \frac{9}<br />
{4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) -  \hfill \\<br />
   - \frac{3}<br />
{{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}<br />
{{\sqrt 7 }}} \right) + C. \hfill \\ <br />
\end{gathered} <br />
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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.

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

    <br />
\begin{gathered}<br />
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 \\<br />
= 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 \\ <br />
\end{gathered} <br />

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

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

    <br />
\Rightarrow \left\{ \begin{gathered}<br />
A = 1, \hfill \\<br />
B + D = 0, \hfill \\<br />
2A + 2B + C = 0, \hfill \\<br />
A - B + C + D = 0; \hfill \\ <br />
\end{gathered} \right.<br />

    <br />
\Leftrightarrow \left\{ \begin{gathered}<br />
A = 1, \hfill \\<br />
B = - D, \hfill \\<br />
2D - C = 2, \hfill \\<br />
2D + C = - 1; \hfill \\ <br />
\end{gathered} \right.<br />

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

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

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

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


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

    <br />
\frac{3}<br />
{4}\int {\frac{{dt}}<br />
{{2t^2 - t + 1}}} = 6\int {\frac{{dt}}<br />
{{16t^2 - 8t + 1 + 7}}} = 6\int {\frac{{dt}}<br />
{{\left( {4t - 1} \right)^2 + 7}}} = <br />

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

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

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