The integral you have presented appears to be:
$\int_1^4 (2 + x^2) x \, dx$

Step-by-step Solution:

1. Expand the integrand $(2 + x^2)x$:
   $(2 + x^2)x = 2x + x^3$

   So the integral becomes:
   $\int_1^4 (2x + x^3) \, dx$

2. Separate the integral:
   $\int_1^4 (2x + x^3) \, dx = \int_1^4 2x \, dx + \int_1^4 x^3 \, dx$

3. Compute each term:

   • For $\int_1^4 2x \, dx$:
     $\int 2x \, dx = x^2$ Evaluate $x^2$ from 1 to 4:
     $[x^2]_1^4 = 4^2 - 1^2 = 16 - 1 = 15$

   • For $\int_1^4 x^3 \, dx$:
     $\int x^3 \, dx = \frac{x^4}{4}$ Evaluate $\frac{x^4}{4}$ from 1 to 4:
     $\left[\frac{x^4}{4}\right]_1^4 = \frac{4^4}{4} - \frac{1^4}{4} = \frac{256}{4} - \frac{1}{4} = 64 - 0.25 = 63.75$

4. Add the results:
   $15 + 63.75 = 78.75$

Final Answer:

$\int_1^4 (2 + x^2)x \, dx = 78.75$

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Related Questions:

1. How can integrals be evaluated using substitution techniques?
2. What are the applications of definite integrals in physics or economics?
3. How does the Fundamental Theorem of Calculus apply to definite integrals?
4. Can this integral be solved graphically or numerically for verification?
5. What is the difference between definite and indefinite integrals?

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Tip:

Always double-check the expansion of polynomials in integrals to ensure correct setup before integrating.