The given integral is:

$\int \sqrt{x^5} \, dx$

Step 1: Rewrite the integrand

The square root can be expressed as a fractional exponent: $\sqrt{x^5} = x^{5/2}$

Thus, the integral becomes: $\int x^{5/2} \, dx$

Step 2: Apply the power rule of integration

The power rule states: \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{(valid if n \neq -1)}.

Here, $n = \frac{5}{2}$, so: $n + 1 = \frac{5}{2} + 1 = \frac{7}{2}$

Thus: $\int x^{5/2} \, dx = \frac{x^{7/2}}{7/2} + C$

Step 3: Simplify the result

Dividing by $\frac{7}{2}$ is equivalent to multiplying by $\frac{2}{7}$, so: $\int x^{5/2} \, dx = \frac{2}{7} x^{7/2} + C$

Final Answer:

$\int \sqrt{x^5} \, dx = \frac{2}{7} x^{7/2} + C$

Would you like a breakdown of any step or further clarifications?

Related Questions:

1. How do you handle integrals with fractional exponents?
2. What is the significance of the constant $C$ in indefinite integrals?
3. How would the integral change if it were $\sqrt[3]{x^5}$?
4. How does this integral relate to definite integrals over a range?
5. What happens if the integrand has a coefficient, like $3\sqrt{x^5}$?

Tip:

When dealing with radicals in integrals, rewriting them in exponential form often simplifies the process.