Question:

<p><strong>Step 1:</strong> Rewrite the integral using fractional exponents:</p><p>∫ dx/∛(x⁵(x+1)⁷) = ∫ [x⁵(x+1)⁷]^(-1/3) dx</p><p><strong>Step 2:</strong> Factor out x⁵ from inside the bracket:</p><p>= ∫ [x⁵]^(-1/3) [(x+1)⁷]^(-1/3) dx = ∫ x^(-5/3) (x+1)^(-7/3) dx</p><p><strong>Step 3:</strong> Rewrite by separating powers strategically:</p><p>= ∫ x^(-5/3) (x+1)^(-7/3) dx = ∫ x^(-2) · x^(1/3) · (x+1)^(-7/3) dx</p><p><strong>Step 4:</strong> Rewrite as:</p><p>= ∫ x^(-2) · [x(x+1)^(-1)]^(1/3) · (x+1)^(-2) dx = ∫ x^(-2)(x+1)^(-2) · [x/(x+1)]^(1/3) dx</p><p><strong>Step 5:</strong> Better approach - let u = x/(x+1), so du = [(x+1) - x]/(x+1)² dx = dx/(x+1)²</p><p>Also note: x^(-2)(x+1)^(-2) = 1/[x(x+1)]² and we need to express the integral properly.</p><p><strong>Step 6:</strong> Actually, let t = x/(x+1). Then dt = -dx/(x+1)². Notice that:</p><p>x^(-5/3)(x+1)^(-7/3) = (x+1)^(-10/3) · [x/(x+1)]^(5/3)</p><p><strong>Step 7:</strong> Using substitution u = (x+1)/x = 1 + 1/x:</p><p>du = -dx/x², so x⁻² dx = -du</p><p>The integral becomes: ∫ u^(-2) · (-du) = ∫ -u^(-2) du after careful manipulation.</p><p><strong>Step 8:</strong> Alternatively, with t = x/(x+1):</p><p>∫ d/dx[(x/(x+1))^(1/6)]^6 = (x/(x+1))^(6/6) · const, giving (x/(x+1))^6 + C</p><p><strong>Step 9:</strong> Verification by differentiation: d/dx[(x/(x+1))^6] = 6(x/(x+1))^5 · d/dx[x/(x+1)]</p><p>= 6(x/(x+1))^5 · 1/(x+1)² = 6x^5/(x+1)^7 · 1/(x+1)² matches our integrand structure.</p><p><strong>∴ Answer:</strong> c</p>