Answer :
- Use u-substitution with $u = x^5 + 3$.
- Rewrite the integral in terms of $u$ as $5 \int (u-3) \sqrt{u} du$.
- Integrate to get $2u^{5/2} - 10u^{3/2} + C$.
- Substitute back to get the final answer: $\boxed{2(x^5+3)^{3/2} (x^5 - 2) + C}$.
### Explanation
1. Problem Analysis
We are asked to evaluate the indefinite integral $\int 25 x^9 \cdot \sqrt{x^5+3} d x$.
2. Applying Substitution
To solve this integral, we will use the substitution method. Let $u = x^5 + 3$. Then, the derivative of $u$ with respect to $x$ is $\frac{du}{dx} = 5x^4$, which means $du = 5x^4 dx$.
3. Rewriting the Integral
Now, we need to rewrite the integral in terms of $u$. Notice that $x^5 = u - 3$. We can rewrite $25x^9 dx$ as $5x^5 \cdot 5x^4 dx = 5(u-3) du$. Therefore, the integral becomes:
$$\int 25 x^9 \sqrt{x^5+3} dx = \int 5(u-3) \sqrt{u} du$$
4. Expanding the Expression
Expanding the expression inside the integral, we get:
$$\int 5(u-3) \sqrt{u} du = 5 \int (u^{3/2} - 3u^{1/2}) du$$
5. Integrating with Respect to u
Now, we integrate with respect to $u$:
$$5 \int (u^{3/2} - 3u^{1/2}) du = 5 \left[ \frac{u^{5/2}}{5/2} - 3 \cdot \frac{u^{3/2}}{3/2} \right] + C = 5 \left[ \frac{2}{5} u^{5/2} - 2 u^{3/2} \right] + C = 2u^{5/2} - 10u^{3/2} + C$$
6. Substituting Back
Substitute back $u = x^5 + 3$ to express the result in terms of $x$:
$$2(x^5+3)^{5/2} - 10(x^5+3)^{3/2} + C$$
7. Simplifying the Expression
We can simplify this expression by factoring out $2(x^5+3)^{3/2}$:
$$2(x^5+3)^{3/2} \left[ (x^5+3) - 5 \right] + C = 2(x^5+3)^{3/2} (x^5 - 2) + C$$
8. Final Answer
Thus, the final result is:
$$\int 25 x^9 \sqrt{x^5+3} dx = 2(x^5+3)^{3/2} (x^5 - 2) + C$$
So, the evaluated integral is $\boxed{2(x^5+3)^{3/2} (x^5 - 2) + C}$.
### Examples
Understanding indefinite integrals is crucial in physics, especially when dealing with motion. For instance, if you know the force acting on an object as a function of its position, you can integrate it to find the potential energy function. This is similar to our problem, where we found an antiderivative; in physics, this antiderivative could represent a physical quantity derived from another related quantity.
- Rewrite the integral in terms of $u$ as $5 \int (u-3) \sqrt{u} du$.
- Integrate to get $2u^{5/2} - 10u^{3/2} + C$.
- Substitute back to get the final answer: $\boxed{2(x^5+3)^{3/2} (x^5 - 2) + C}$.
### Explanation
1. Problem Analysis
We are asked to evaluate the indefinite integral $\int 25 x^9 \cdot \sqrt{x^5+3} d x$.
2. Applying Substitution
To solve this integral, we will use the substitution method. Let $u = x^5 + 3$. Then, the derivative of $u$ with respect to $x$ is $\frac{du}{dx} = 5x^4$, which means $du = 5x^4 dx$.
3. Rewriting the Integral
Now, we need to rewrite the integral in terms of $u$. Notice that $x^5 = u - 3$. We can rewrite $25x^9 dx$ as $5x^5 \cdot 5x^4 dx = 5(u-3) du$. Therefore, the integral becomes:
$$\int 25 x^9 \sqrt{x^5+3} dx = \int 5(u-3) \sqrt{u} du$$
4. Expanding the Expression
Expanding the expression inside the integral, we get:
$$\int 5(u-3) \sqrt{u} du = 5 \int (u^{3/2} - 3u^{1/2}) du$$
5. Integrating with Respect to u
Now, we integrate with respect to $u$:
$$5 \int (u^{3/2} - 3u^{1/2}) du = 5 \left[ \frac{u^{5/2}}{5/2} - 3 \cdot \frac{u^{3/2}}{3/2} \right] + C = 5 \left[ \frac{2}{5} u^{5/2} - 2 u^{3/2} \right] + C = 2u^{5/2} - 10u^{3/2} + C$$
6. Substituting Back
Substitute back $u = x^5 + 3$ to express the result in terms of $x$:
$$2(x^5+3)^{5/2} - 10(x^5+3)^{3/2} + C$$
7. Simplifying the Expression
We can simplify this expression by factoring out $2(x^5+3)^{3/2}$:
$$2(x^5+3)^{3/2} \left[ (x^5+3) - 5 \right] + C = 2(x^5+3)^{3/2} (x^5 - 2) + C$$
8. Final Answer
Thus, the final result is:
$$\int 25 x^9 \sqrt{x^5+3} dx = 2(x^5+3)^{3/2} (x^5 - 2) + C$$
So, the evaluated integral is $\boxed{2(x^5+3)^{3/2} (x^5 - 2) + C}$.
### Examples
Understanding indefinite integrals is crucial in physics, especially when dealing with motion. For instance, if you know the force acting on an object as a function of its position, you can integrate it to find the potential energy function. This is similar to our problem, where we found an antiderivative; in physics, this antiderivative could represent a physical quantity derived from another related quantity.
