To find $f(x)$ given its derivative $f'(x)$ and the condition $f(1) = 1$, we need to integrate $f'(x)$ and then use the condition to solve for the constant of integration.

Step-by-Step Solution:

1. Rewrite $f'(x)$: $f'(x) = 4x^3 + 2x - 5x^9 + 6x^2 + 10x^4 - 6x^{-2} + \frac{4}{3}x^{-\frac{1}{3}} - 12$

2. Integrate $f'(x)$ term-by-term:

   • For $4x^3$: $\int 4x^3 \, dx = x^4$

   • For $2x$: $\int 2x \, dx = x^2$

   • For $-5x^9$: $\int -5x^9 \, dx = -\frac{5}{10}x^{10} = -\frac{1}{2}x^{10}$

   • For $6x^2$: $\int 6x^2 \, dx = 2x^3$

   • For $10x^4$: $\int 10x^4 \, dx = 2x^5$

   • For $-6x^{-2}$: $\int -6x^{-2} \, dx = \frac{6}{x}$

   • For $\frac{4}{3}x^{-\frac{1}{3}}$: $\int \frac{4}{3}x^{-\frac{1}{3}} \, dx = \frac{4}{3} \cdot \frac{3}{2}x^{\frac{2}{3}} = 2x^{\frac{2}{3}}$

   • For $-12$: $\int -12 \, dx = -12x$

   Now, summing up all these results: $f(x) = x^4 + x^2 - \frac{1}{2}x^{10} + 2x^3 + 2x^5 + \frac{6}{x} + 2x^{\frac{2}{3}} - 12x + C$

3. Use the Initial Condition $f(1) = 1$:

   Substitute $x = 1$ and $f(1) = 1$ into $f(x)$: $1^4 + 1^2 - \frac{1}{2}(1)^{10} + 2(1)^3 + 2(1)^5 + \frac{6}{1} + 2(1)^{\frac{2}{3}} - 12(1) + C = 1$ Simplify: $1 + 1 - \frac{1}{2} + 2 + 2 + 6 + 2 - 12 + C = 1$ $2 + 1.5 + 2 - 12 + C = 1$ $4.5 - 12 + C = 1$ $C = 8.5$

Final Answer:

$f(x) = x^4 + x^2 - \frac{1}{2}x^{10} + 2x^3 + 2x^5 + \frac{6}{x} + 2x^{\frac{2}{3}} - 12x + 8.5$

Would you like further explanation on the integration steps or any other details?

---

Related Questions:

1. How do we approach integrating more complex functions such as logarithmic or trigonometric derivatives?
2. What are common mistakes to avoid when integrating powers of $x$?
3. How does one integrate terms involving fractional exponents?
4. What are some applications of indefinite integrals in physics?
5. How does one find an antiderivative when a derivative contains products of functions?

Tip:

When integrating, always double-check the power and the constant factors to ensure you are applying the power rule correctly for each term.