<p>Evaluate: <span class="math">\int x^5 \sqrt{1+x^3} dx</span></p>
<p>(a) <span class="math">\frac{1}{15}(1+x^3)^{5/2} - \frac{1}{9}(1+x^3)^{3/2} + c</span></p>
<p>(b) <span class="math">\frac{2}{15}(1+x^3)^{5/2} - \frac{1}{9}(1+x^3)^{3/2} + c</span></p>
<p>(c) <span class="math">\frac{2}{15}(1+x^3)^{5/2} - \frac{2}{9}(1+x^3)^{3/2} + c</span></p>
<p>(d) <span class="math">\frac{1}{15}(1+x^3)^{5/2} - \frac{2}{9}(1+x^3)^{3/2} + c</span></p>
Step-by-Step Solution
Key Concept: Recognize that x^5 can be rewritten in terms of powers of x^3, and use substitution u = 1 + x^3 to transform the integral into a polynomial form in u that can be integrated directly.
<p><strong>Step 1:</strong> Rewrite the integrand by expressing x^5 in a useful form.</p><p>$$\int x^5\sqrt{1+x^3}\,dx = \int x^3 \cdot x^2 \sqrt{1+x^3}\,dx$$</p><p><strong>Step 2:</strong> Use substitution u = 1 + x³, so du = 3x² dx, which gives x² dx = du/3.</p><p>Also, from u = 1 + x³, we have x³ = u - 1.</p><p><strong>Step 3:</strong> Substitute into the integral:</p><p>$$\int x^5\sqrt{1+x^3}\,dx = \int (u-1)\sqrt{u} \cdot \frac{du}{3}$$</p><p>$$= \frac{1}{3}\int (u-1)u^{1/2}\,du$$</p><p><strong>Step 4:</strong> Expand the integrand:</p><p>$$= \frac{1}{3}\int (u^{3/2} - u^{1/2})\,du$$</p><p><strong>Step 5:</strong> Integrate term by term using the power rule:</p><p>$$= \frac{1}{3}\left[\frac{u^{5/2}}{5/2} - \frac{u^{3/2}}{3/2}\right] + c$$</p><p>$$= \frac{1}{3}\left[\frac{2u^{5/2}}{5} - \frac{2u^{3/2}}{3}\right] + c$$</p><p><strong>Step 6:</strong> Simplify the coefficients:</p><p>$$= \frac{2u^{5/2}}{15} - \frac{2u^{3/2}}{9} + c$$</p><p><strong>Step 7:</strong> Substitute back u = 1 + x³:</p><p>$$= \frac{2}{15}(1+x^3)^{5/2} - \frac{2}{9}(1+x^3)^{3/2} + c$$</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c