Answer on Question #51629 – Math – Integral Calculus
What is the integration of 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) \frac{1}{\sqrt{(x - 1)(x - 2)(x - 3)}} ( x − 1 ) ( x − 2 ) ( x − 3 ) 1
Solution
Let
I ( x ) = ∫ 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) d x . I(x) = \int \frac{1}{\sqrt{(x - 1)(x - 2)(x - 3)}} \, dx. I ( x ) = ∫ ( x − 1 ) ( x − 2 ) ( x − 3 ) 1 d x .
This integral cannot be expressed in elementary functions.
We use Wolfram Mathematica online integrator and find
I ( x ) = ( 2 i ( 1 x − 3 + 1 ) ( 2 x − 3 + 1 ) ( x − 3 ) 3 2 ⋅ F ( i sinh − 1 ( 1 x − 3 ) ∣ 2 ) ) ( x − 1 ) ( x − 2 ) ( x − 3 ) , I(x) = \frac{\left(2i\sqrt{\left(\frac{1}{x - 3} + 1\right)} \sqrt{\left(\frac{2}{x - 3} + 1\right)} (x - 3)^{\frac{3}{2}} \cdot F\left(i \sinh^{-1}\left(\frac{1}{\sqrt{x - 3}}\right) \mid 2\right)\right)}{\sqrt{(x - 1)(x - 2)(x - 3)}}, I ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( 2 i ( x − 3 1 + 1 ) ( x − 3 2 + 1 ) ( x − 3 ) 2 3 ⋅ F ( i sinh − 1 ( x − 3 1 ) ∣ 2 ) ) ,
where i i i is imaginary unit, sinh − 1 ( x ) \sinh^{-1}(x) sinh − 1 ( x ) is inverse hyperbolic sine, F ( x ∣ m ) F(x|m) F ( x ∣ m ) is elliptic integral of the first kind.
We can simplify it to
I ( x ) = ( 2 i ( x − 2 x − 3 ) ( x − 1 x − 3 ) ( x − 3 ) 3 2 ⋅ F ( i sinh − 1 ( 1 x − 3 ) ∣ 2 ) ) ( x − 1 ) ( x − 2 ) ( x − 3 ) = 2 i ⋅ F ( i sinh − 1 ( 1 x − 3 ) ∣ 2 ) . I(x) = \frac{\left(2i\sqrt{\left(\frac{x - 2}{x - 3}\right)} \sqrt{\left(\frac{x - 1}{x - 3}\right)} (x - 3)^{\frac{3}{2}} \cdot F\left(i \sinh^{-1}\left(\frac{1}{\sqrt{x - 3}}\right) \mid 2\right)\right)}{\sqrt{(x - 1)(x - 2)(x - 3)}} = 2i \cdot F\left(i \sinh^{-1}\left(\frac{1}{\sqrt{x - 3}}\right) \mid 2\right). I ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( 2 i ( x − 3 x − 2 ) ( x − 3 x − 1 ) ( x − 3 ) 2 3 ⋅ F ( i sinh − 1 ( x − 3 1 ) ∣ 2 ) ) = 2 i ⋅ F ( i sinh − 1 ( x − 3 1 ) ∣ 2 ) .
Answer: 2 i ⋅ F ( i sinh − 1 ( 1 x − 3 ) ∣ 2 ) 2i \cdot F\left(i \sinh^{-1}\left(\frac{1}{\sqrt{x - 3}}\right) \mid 2\right) 2 i ⋅ F ( i sinh − 1 ( x − 3 1 ) ∣ 2 ) .
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