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一个从Mathematica中发现的等式

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青青子衿 Posted 2014-6-27 19:48 |Read mode
  1. f[x]:=Sqrt[-x^2+2x]
  2. \[Integral]f[x] \[DifferentialD]x
  3. F[x] := \[Integral]f[x] \[DifferentialD]x
  4. \!\(\*SubscriptBox[\(\[PartialD]\), \(x\)]\(F[x]\)\)
  5. FullSimplify[\!\(\*SubscriptBox[\(\[PartialD]\), \(x\)]\(F[x]\)\)]
  6. F2[x] := 1/2 ArcSin[x - 1] + 1/2 (x - 1) Sqrt[2 x - x^2]
  7. \!\(\*SubscriptBox[\(\[PartialD]\), \(x\)]\(F2[x]\)\)
  8. Simplify[\!\(\*SubscriptBox[\(\[PartialD]\), \(x\)]\(F2[x]\)\)]
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\[f\left( x \right) = \sqrt {x\left( {2 - x} \right)}\]

\[\int {f\left( x \right)dx}\]

\[\begin{gathered}
  F\left( x \right) = \int {f\left( x \right)dx} \\
   = \frac{{\sqrt {x\left( {2 - x} \right)} \left( {\left( {x - 1} \right)\sqrt {x\left( {x - 2} \right)}  - 2\log \left[ {\sqrt {x - 2}  + \sqrt x } \right]} \right)}}{{2\sqrt {x\left( {x - 2} \right)} }} \\
   = \log \left[ {\sqrt {x - 2}  + \sqrt x } \right] - \frac{{\left( {x - 1} \right)\sqrt {x\left( {x - 2} \right)} }}{2} \\
\end{gathered}\]

\[\begin{gathered}
  F\prime \left( x \right) \\
   = \frac{{\left( { - \frac{{\frac{1}{{\sqrt {x - 2} }} + \frac{1}{{\sqrt x }}}}{{\sqrt {x - 2}  + \sqrt x }} + \frac{{\left( {x - 1} \right)}}{2}\sqrt {\frac{{x - 2}}{x}}  + \sqrt {x\left( {x - 2} \right)}  + \frac{{\left( {x - 1} \right)}}{2}\sqrt {\frac{x}{{x - 2}}} } \right)\sqrt {x\left( {2 - x} \right)} }}{{2\sqrt {x\left( {x - 2} \right)} }} +  \\
  \frac{{\left( {2 - 2x} \right)\left( {\left( {x - 1} \right)\sqrt {x\left( {x - 2} \right)}  - 2\log \left[ {\sqrt {x - 2}  + \sqrt x } \right]} \right)}}{{4\sqrt {x\left( {x - 2} \right)} \sqrt {x\left( {2 - x} \right)} }} -  \\
  \frac{{\sqrt {x\left( {2 - x} \right)} \left( {\left( {x - 1} \right)\sqrt {x\left( {x - 2} \right)}  - 2\log \left[ {\sqrt {x - 2}  + \sqrt x } \right]} \right)}}{{4{x^{3/2}}\sqrt {x - 2} }} -  \\
  \frac{{\sqrt {x\left( {2 - x} \right)} \left( {\left( {x - 1} \right)\sqrt {x\left( {x - 2} \right)}  - 2\log \left[ {\sqrt {x - 2}  + \sqrt x } \right]} \right)}}{{4{{\left( {x - 2} \right)}^{3/2}}\sqrt x }} \\
\end{gathered} \]

\[Simplify\left[ {F\prime \left( x \right)} \right] = \sqrt {2x - {x^2}} \]

\[{F_2}\left( x \right) = \arcsin \left( {x - 1} \right) + \left( {x - 1} \right)\sqrt {2x - {x^2}} \]

\[{F_2}\prime \left( x \right) = \frac{1}{{2\sqrt {1 - {{\left( {1 - x} \right)}^2}} }} + \frac{{\left( {2 - 2x} \right)\left( {x - 1} \right)}}{{4\sqrt {2x - {x^2}} }} + \frac{1}{2}\sqrt {2x - {x^2}} \]

\[Simplify\left[ {{F_2}\prime \left( x \right)} \right] = \sqrt {2x - {x^2}} \]
由\[F(x)=F_2(x)\]得:
\[\begin{gathered}
\color{red}{\arcsin \left( {x - 1} \right) + \left( {x - 1} \right)\sqrt {x\left( {2 - x} \right)} \\
   = \log \left[ {\sqrt {x - 2}  + \sqrt x } \right] - \frac{{\left( {x - 1} \right)\sqrt {x\left( {x - 2} \right)} }}{2}} \\
\end{gathered} \]

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hbghlyj Posted 2023-3-26 17:17
$\arcsin(x-1)$需要$x-1≤1$
$\sqrt{x-2}$需要$x≥2$

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2025-6-4 17:31 GMT+8

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