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Step-by-Step Solution
Step 1: Identify the Functions f(x) and g(x)
We have two functions defined for $x > 0$:
$f(x) = \displaystyle \int_{-x}^{x} \bigl(|t| - t^2\bigr)\,e^{-t^2}\,dt,$
and
$g(x) = \displaystyle \int_{0}^{x^2} t^{\tfrac{1}{2}}\,e^{-t}\,dt.$
Our goal is to evaluate
$9\bigl(f(\sqrt{\log_e 9}) + g(\sqrt{\log_e 9})\bigr).$
Step 2: Differentiate f(x)
Consider
$f(x) = \displaystyle \int_{-x}^{x} (|t| - t^2)\,e^{-t^2}\,dt.$
For $x>0$, $|x| = x$. Using the Leibniz rule for the derivative of an integral with variable limits:
$$
f'(x)
= \Bigl((|x| - x^2)e^{-x^2}\Bigr)\cdot \frac{d}{dx}(x)
\;-\; \Bigl((|-x| - (-x)^2)e^{-(-x)^2}\Bigr)\cdot \frac{d}{dx}(-x).
$$
Since $|x| = x$ (for $x > 0$), $\frac{d}{dx}(x) = 1$, and $\frac{d}{dx}(-x) = -1$, we get:
$$
f'(x)
= \bigl(x - x^2\bigr)e^{-x^2}
\;-\;
\bigl(x - x^2\bigr)e^{-x^2} \times(-1)
= 2\,\bigl(x - x^2\bigr)\,e^{-x^2}.
$$
Step 3: Differentiate g(x)
Consider
$$
g(x) = \int_{0}^{x^2} t^{\tfrac{1}{2}}\, e^{-t}\,dt.
$$
By the Fundamental Theorem of Calculus and the chain rule:
$$
g'(x)
= (x^2)^{\tfrac{1}{2}}\;e^{-\,x^2}
\times \frac{d}{dx}(x^2)
= x \, e^{-\,x^2} \times 2x
= 2x^2 \, e^{-\,x^2}.
$$
Step 4: Sum the Derivatives f'(x) + g'(x)
Adding the two derivatives:
$$
f'(x) + g'(x)
= 2\,(x - x^2)\, e^{-x^2} + 2x^2\, e^{-x^2}
= 2x\, e^{-x^2}.
$$
Step 5: Integrate to Find f(x) + g(x)
Since
$$
(f + g)'(x) = 2x\, e^{-\,x^2},
$$
we integrate from 0 to $x$:
$$
(f + g)(x) - (f + g)(0)
= \int_{0}^{x} 2u\, e^{-u^2}\,du.
$$
Notice that $f(0) = 0$ and $g(0) = 0$, so $(f+g)(0) = 0$. Thus:
$$
(f + g)(x)
= \int_{0}^{x} 2u\, e^{-u^2}\,du.
$$
The integral
$\displaystyle \int 2u\, e^{-u^2}\,du = -\, e^{-u^2}.$
Hence,
$$
(f + g)(x)
= \Bigl[-\,e^{-u^2}\Bigr]_{0}^{\,x}
= \bigl(-\,e^{-\,x^2}\bigr)\;-\;\bigl(-\,e^{0}\bigr)
= 1 \;-\; e^{-\,x^2}.
$$
Step 6: Evaluate at $x = \sqrt{\log_e(9)}$
Let $x_0 = \sqrt{\log_e(9)}.$ Then
$$
(f + g)(x_0)
= 1 - e^{-\,x_0^2}
= 1 - e^{-\,\log_e(9)}.
$$
Since $e^{-\,\log_e(9)} = \tfrac{1}{e^{\log_e(9)}} = \tfrac{1}{9},$ we have:
$$
(f + g)\bigl(\sqrt{\log_e(9)}\bigr)
= 1 - \tfrac{1}{9}
= \tfrac{8}{9}.
$$
Step 7: Multiply by 9 to Get the Final Answer
Therefore,
$$
9\bigl(f(\sqrt{\log_e(9)}) + g(\sqrt{\log_e(9)})\bigr)
= 9 \times \frac{8}{9} = 8.
$$
Hence, the value is $\boxed{8}.$
