For the function $$ f $$, given below, find the antiderivative $$ F $$ that satisfies $$ F(1)=1 $$. $$ f(x)=x^{6}-3 x^{-4}+5 $$ The antiderivative that satisfies the given condition is $$ F(x)=\square $$.

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To find the antiderivative $$ F(x) $$ of the function $$ f(x) = x^6 - 3x^{-4} + 5 $$ that satisfies $$ F(1) = 1 $$, follow these steps:

1. **Integrate Each Term Individually:**

$$
   \begin{align*}
   \int x^6 \, dx &= \frac{x^7}{7} \
   \int -3x^{-4} \, dx &= -3 \cdot \left( \frac{x^{-3}}{-3} \right) = x^{-3} \
   \int 5 \, dx &= 5x \
   \end{align*}
   $$

2. **Combine the Integrated Terms and Add the Constant of Integration $$ C $$:**

$$
   F(x) = \frac{x^7}{7} + \frac{1}{x^3} + 5x + C
   $$

3. **Apply the Initial Condition $$ F(1) = 1 $$ to Determine $$ C $$:**

$$
   \begin{align*}
   F(1) &= \frac{1^7}{7} + \frac{1}{1^3} + 5(1) + C = \frac{1}{7} + 1 + 5 + C = \frac{43}{7} + C = 1 \
   \Rightarrow C &= 1 - \frac{43}{7} = -\frac{36}{7}
   \end{align*}
   $$

4. **Write the Final Expression for $$ F(x) $$:**

$$
   F(x) = \frac{x^7}{7} + \frac{1}{x^3} + 5x - \frac{36}{7}
   $$

**Final Answer:**
$$ F(x) = \frac{x^{7}}{7} + \frac{1}{x^{3}} + 5\,x\ -\ \frac{36}{7} $$
The antiderivative $$ F(x) $$ is $$ \frac{x^{7}}{7} + \frac{1}{x^{3}} + 5x - \frac{36}{7} $$.

For the function \( f \), given below, find the antiderivative \( F \) that satisfies \( F(1)=1 \). \( f(x)=x^{6}-3 x^{-4}+5 \) The antiderivative that satisfies the given condition is \( F(x)=\square \).

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