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| Indefinite Integral http://mathhelpplanet.com/viewtopic.php?f=19&t=16345 |
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| Автор: | jagdish [ 22 апр 2012, 21:56 ] |
| Заголовок сообщения: | Indefinite Integral |
[math]\displaystyle \int \frac{x^3-2}{\sqrt{(x^3+1)^2}}dx[/math] |
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| Автор: | erjoma [ 22 апр 2012, 22:33 ] |
| Заголовок сообщения: | Re: Indefinite Integral |
[math]\begin{gathered} \int {\frac{{{x^3} - 2}}{{\sqrt {{{\left( {{x^3} + 1} \right)}^2}} }}dx} = \int {\frac{{{x^3} - 2}}{{\left| {{x^3} + 1} \right|}}dx} = \int {\operatorname{sign} \left( {x + 1} \right)} \frac{{{x^3} - 2}}{{{x^3} + 1}}dx = \operatorname{sign} \left( {x + 1} \right)\int {\left( {1 - \frac{3}{{{x^3} + 1}}} \right)dx} = ... \hfill \\ \end{gathered}[/math] P.S. [math]\operatorname{sign} \left( x \right) = \left\{ \begin{gathered} 1,x > 0 \hfill \\ 0,x = 0 \hfill \\ - 1,x < 0 \hfill \\ \end{gathered} \right.[/math] |
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| Автор: | jagdish [ 23 апр 2012, 04:58 ] |
| Заголовок сообщения: | Re: Indefinite Integral |
thanks arjoma,actually original question is [math]\displaystyle \int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx[/math] |
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| Автор: | Prokop [ 23 апр 2012, 08:46 ] |
| Заголовок сообщения: | Re: Indefinite Integral |
The answer is found by accident. The solution is based on the identity [math]\int {\frac{{x^3 - 2}}{{\sqrt {\left( {x^3 + 1} \right)^3 } }}dx} = \frac{{P\left( x \right)}}{{\sqrt {x^3 + 1} }} + C[/math] [math]P\left( x \right)[/math] - polynomial of degree 1 The answer [math]\frac{{ - 2x}}{{\sqrt {x^3 + 1} }} + C[/math] |
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| Автор: | jagdish [ 23 апр 2012, 09:13 ] |
| Заголовок сообщения: | Re: Indefinite Integral |
Thanks Prokop very Nice solution. any other Method. |
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| Автор: | Prokop [ 23 апр 2012, 10:43 ] |
| Заголовок сообщения: | Re: Indefinite Integral |
Another method is based on the theory of binomial differentials [math]J_{p,q} = \int {\left( {a + bt} \right)^p t^q dt}[/math] There is equality [math]J_{p,q} = \frac{{\left( {a + bt} \right)^{p + 1} t^{q + 1} }}{{a\left( {q + 1} \right)}} - b\frac{{p + q + 2}}{{a\left( {q + 1} \right)}}J_{p,q + 1}[/math] Your integral is equal to [math]I=\int {\frac{{x^3 - 2}}{{\sqrt {\left( {x^3 + 1} \right)^3 } }}dx}= \left\{ {x^3 = t} \right\} = \frac{1}{3}\left( {J_{ - \frac{3}{2},\frac{1}{3}} - 2J_{ - \frac{3}{2}, - \frac{2}{3}} } \right)[/math] here [math]J_{p,q} = \int {\left( {1 + t} \right)^p t^q dt}[/math] Further [math]J_{ - \frac{3}{2}, - \frac{2}{3}}= \frac{{\left({1 + t} \right)^{ - \frac{1}{2}} t^{\frac{1}{3}} }}{{\frac{1}{3}}} - \frac{{ - \frac{3}{2} - \frac{2}{3} + 2}}{{\frac{1}{3}}}J_{ - \frac{3}{2},\frac{1}{3}}[/math] Therefore [math]I = \frac{{ - 2x}}{{\sqrt {x^3 + 1} }} + C[/math] |
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