Integrate the function 1 8 + 3x - x^2

Question

Integrate the function $\frac{1}{{\sqrt {8 + 3x - {x^2}} }}$

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Answer

$\int {\frac{1}{{\sqrt {8 + 3x - {x^2}} }}} dx$ $ = \int {\frac{1}{{\sqrt { - {x^2} + 3x + 8} }}} dx$
$= \int {\frac{1}{{\sqrt { - \left( {{x^2} - 3x - 8} \right)} }}} dx$
$= \int {\frac{1}{{\sqrt { - \left\{ {{x^2} - 3x + {{\left( {\frac{3}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2} - 8} \right\}} }}dx} $
$ = \int {\frac{1}{{\sqrt { - \left\{ {{{\left( {x - \frac{3}{2}} \right)}^2} - \frac{{41}}{4}} \right\}} }}dx} $
$= \int {\frac{1}{{\sqrt {{{\left( {\frac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \frac{3}{2}} \right)}^2}} }}dx} $
$= {\sin ^{ - 1}}\frac{{x - \frac{3}{2}}}{{\frac{{\sqrt {41} }}{2}}} + c$
$= {\sin ^{ - 1}}\left( {\frac{{2x - 3}}{{\sqrt {41} }}} \right) + c$

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