Integral. Verification of decision

• Dec 31st 2008, 05:14 PM
DeMath
Integral. Verification of decision
Can I find this integral by more simple method?
I think that I used bad replacement of the variable of integration.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \begin{gathered} \boxed{\int {\frac{{dx}} {{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} } = \ln x - \frac{3} {2}\ln \left( {\sqrt[6]{x} + 1} \right) - \frac{9} {4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) - \hfill \\ - \frac{3} {{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}} {{\sqrt 7 }}} \right) + C. \hfill \\ \end{gathered}$
• Dec 31st 2008, 06:13 PM
NonCommAlg
Quote:

Originally Posted by DeMath
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!