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The integral presented is:

\[
\int \frac{x^3}{\sqrt{x^2 + 5}} \, dx
\]

### Steps to solve:

1. **Substitution**: Use \( u = x^2 + 5 \), which simplifies the square root.
   - Then, \( du = 2x \, dx \), or \( \frac{du}{2} = x \, dx \).

2. Rewriting \( x^3 \) in terms of \( u \):
   - Notice that \( x^3 = x \cdot x^2 = x \cdot (u - 5) \), so:
   \[
   x^3 = x(u - 5)
   \]

3. Substituting into the integral:
   \[
   \int \frac{x^3}{\sqrt{x^2 + 5}} \, dx = \int \frac{x(u - 5)}{\sqrt{u}} \cdot \frac{du}{2}
   \]

4. Breaking it into two simpler integrals:
   \[
   = \frac{1}{2} \int (u - 5) u^{-\frac{1}{2}} \, du
   \]
   \[
   = \frac{1}{2} \left( \int u^{\frac{1}{2}} \, du - 5 \int u^{-\frac{1}{2}} \, du \right)
   \]

5. Solving these integrals:
   - \( \int u^{\frac{1}{2}} \, du = \frac{2}{3} u^{\frac{3}{2}} \)
   - \( \int u^{-\frac{1}{2}} \, du = 2 u^{\frac{1}{2}} \)

6. Putting everything together:
   \[
   \frac{1}{2} \left( \frac{2}{3} u^{\frac{3}{2}} - 5 \cdot 2 u^{\frac{1}{2}} \right)
   \]
   \[
   = \frac{1}{3} u^{\frac{3}{2}} - 5 u^{\frac{1}{2}}
   \]

7. Substitute back \( u = x^2 + 5 \):
   \[
   = \frac{1}{3} (x^2 + 5)^{\frac{3}{2}} - 5 (x^2 + 5)^{\frac{1}{2}} + C
   \]

Thus, the solution to the integral is:

\[
\frac{1}{3} (x^2 + 5)^{\frac{3}{2}} - 5 (x^2 + 5)^{\frac{1}{2}} + C
\]

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. How does substitution simplify integrals involving square roots?
2. What are the common techniques for solving integrals of this type?
3. Can you apply similar substitution methods to trigonometric integrals?
4. What is the geometric interpretation of integrals like this one?
5. How does the choice of substitution affect the simplicity of the solution?

**Tip**: When dealing with square roots and powers of variables in integrals, substitution is a powerful tool for simplification!

Evaluate the integral ∫ (x^3 / √(x^2 + 5)) dx.

Learn how to evaluate the integral ∫ (x^3 / √(x^2 + 5)) dx using substitution. This solution involves simplifying the expression through u-substitution and integrating powers, providing a step-by-step approach suitable for college-level calculus students.

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