Question

The integral \( \int \frac{x^5}{\sqrt{1+x^3}} \, dx \) is equal to:

• A. \( \frac{2}{3}(1+x^3)^{3/2} + \frac{1}{3}(1+x^3)^{1/2} + c \)
• B. \( \frac{2}{9}(1+x^3)^{1/2} - \frac{2}{3}(1+x^3)^{3/2} + c \)
• C. \( \frac{2}{3}(1+x^3)^{3/2} - \frac{1}{3}(1+x^3)^{1/2} + c \)
• D. \( \frac{2}{9}(1+x^3)^{3/2} - \frac{2}{3}(1+x^3)^{1/2} + c \)

Hint:

When the numerator has a higher power of \( x \) than the differential part, express the whole numerator in terms of the substitution variable.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use the substitution method to simplify the integral. Let \( u = 1+x^3 \), then \( du = 3x^2 dx \).

Step 2: Key Formula or Approach:
Rewrite the integral as \( \int \frac{x^3 \cdot x^2}{\sqrt{1+x^3}} \, dx \).
Since \( u = 1+x^3 \), we have \( x^3 = u-1 \).

Step 3: Detailed Explanation:
Substitute \( x^3 = u-1 \) and \( x^2 dx = du/3 \):
\[ \int \frac{u-1}{\sqrt{u}} \cdot \frac{du}{3} = \frac{1}{3} \int (u^{1/2} - u^{-1/2}) \, du \]
\[ = \frac{1}{3} \left( \frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2} \right) + c = \frac{1}{3} \left( \frac{2}{3}u^{3/2} - 2u^{1/2} \right) + c \]
\[ = \frac{2}{9}u^{3/2} - \frac{2}{3}u^{1/2} + c \]
Substituting \( u = 1+x^3 \):
\[ = \frac{2}{9}(1+x^3)^{3/2} - \frac{2}{3}(1+x^3)^{1/2} + c \]

Step 4: Final Answer:
The result is option (D).