Question

\( \int \frac{x^5}{\sqrt{1+x^3}} \, dx = \)

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

Hint:

Always split powers to match substitution derivative.

Answer 1

The Correct Option is D

Concept: Use substitution method when expression inside root has derivative present in numerator.

Step 1: Choose substitution.
Let: \[ u = 1 + x^3 \]

Step 2: Differentiate substitution.
\[ \frac{du}{dx} = 3x^2 \] \[ du = 3x^2 dx \] \[ x^2 dx = \frac{du}{3} \]

Step 3: Rewrite the integral.
Given: \[ \int \frac{x^5}{\sqrt{1+x^3}} dx \] Write: \[ x^5 = x^3 \cdot x^2 \] So: \[ \int \frac{x^3 \cdot x^2}{\sqrt{1+x^3}} dx \]

Step 4: Substitute values.
\[ x^3 = u - 1 \] \[ x^2 dx = \frac{du}{3} \] Thus integral becomes: \[ \int \frac{(u-1)}{\sqrt{u}} \cdot \frac{du}{3} \] \[ = \frac{1}{3} \int \frac{u-1}{\sqrt{u}} du \]

Step 5: Simplify integrand.
\[ \frac{u-1}{\sqrt{u}} = \frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}} \] \[ = u^{1/2} - u^{-1/2} \]

Step 6: Integrate term-by-term.
\[ \frac{1}{3} \int (u^{1/2} - u^{-1/2}) du \] \[ = \frac{1}{3} \left[ \frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2} \right] \]

Step 7: Simplify coefficients.
\[ = \frac{1}{3} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right] \] \[ = \frac{2}{9}u^{3/2} - \frac{2}{3}u^{1/2} \]

Step 8: Take common factor.
\[ = \frac{2}{9}u^{1/2}(u - 3) \]

Step 9: Back substitute \(u = 1 + x^3\).
\[ = \frac{2}{9}\sqrt{1+x^3}[(1+x^3) - 3] \] \[ = \frac{2}{9}\sqrt{1+x^3}(x^3 - 2) \]

Step 10: Final Answer.
\[ \boxed{\frac{2}{9}\sqrt{1+x^3}(x^3 - 2) + C} \]