Question
Integrate the function $\frac{1}{{\sqrt {7 - 6x - {x^2}} }}$

Answer

$\int {\frac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} $ $ = \int {\frac{1}{{\sqrt { - {x^2} - 6x + 7} }}dx} $
$= \int {\frac{1}{{\sqrt { - \left( {{x^2} + 6x - 7} \right)} }}dx} $
$= \int {\frac{1}{{\sqrt { - \left( {{x^2} + 6x + 9 - 9 - 7} \right)} }}dx} $
$= \int {\frac{1}{{\sqrt { - \left\{ {{{\left( {x + 3} \right)}^2} - 16} \right\}} }}dx} $
$= \int {\frac{1}{{\sqrt {{{\left( 16 \right)}} - {{\left( {x + 3} \right)}^2}} }}dx} $
$= \int {\frac{1}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( {x + 3} \right)}^2}} }}dx} $
$= {\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + c$   ...$\left[ {\because \int {\frac{1}{{{a^2} - {x^2}}}dx = {{\sin }^{ - 1}}\frac{x}{a}} } \right]$

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