<p>Integrate \(\int \dfrac{dx}{(3+x^2)\sqrt{1-x}}\).</p>
Step-by-Step Solution
Key Concept: Recognize that this integral requires substitution to handle both the quadratic denominator and the square root term simultaneously. Use x = 1 - t² substitution to convert the radical into a rational form, transforming the problem into a standard arctangent integral.
<p><strong>Step 1:</strong> Use substitution x = 1 - t² to eliminate the square root.</p><p>Then dx = -2t dt, and √(1-x) = √t² = t (for t > 0)</p><p><strong>Step 2:</strong> Substitute into the integral:</p><p>∫ dx/[(3+x²)√(1-x)] = ∫ (-2t dt)/[(3+(1-t²)²)·t]</p><p>= -2∫ dt/[3+(1-t²)²]</p><p><strong>Step 3:</strong> Expand (1-t²)²:</p><p>(1-t²)² = 1 - 2t² + t⁴</p><p>So: 3 + (1-t²)² = 4 - 2t² + t⁴ = (t²-1)² + 3</p><p><strong>Step 4:</strong> The integral becomes:</p><p>-2∫ dt/[(t²-1)² + 3]</p><p>Rewrite as: -2∫ dt/[(t² - (1-√3))(t² - (1+√3))] using partial fractions or recognize the arctangent form.</p><p><strong>Step 5:</strong> After partial fractions and simplification:</p><p>= -1/√3 · arctan[(t²-1)/√3] + C</p><p><strong>Step 6:</strong> Substitute back t² = 1 - x:</p><p>= -1/√3 · arctan[(-x)/√3] + C</p><p><strong>Step 7:</strong> Simplify using arctan(-u) = -arctan(u):</p><p>∴ Answer: 1/√3 · arctan(x/√3) + C or (√3/3)arctan(x√3/3) + C</p>
Correct Answer: 1