Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Evaluate \(\int \dfrac{dx}{(x^2+1)\sqrt{x}}\).</p>
<p>\(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x-1}{\sqrt{2x}}\right) - \dfrac{1}{2\sqrt{2}}\log\left|\dfrac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right| + c\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x-1}{\sqrt{2x}}\right) + \dfrac{1}{2\sqrt{2}}\log\left|\dfrac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right| + c\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x-1}{\sqrt{2x}}\right) - \dfrac{1}{\sqrt{2}}\log\left|\dfrac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right| + c\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Use the substitution √x = tan(θ) to convert the integral into a standard trigonometric form, which simplifies the denominator structure and reveals a recognizable antiderivative pattern.
<p><strong>Step 1:</strong> Let √x = tan(θ), so x = tan²(θ), dx = 2tan(θ)sec²(θ)dθ</p><p><strong>Step 2:</strong> Substitute into the integral:</p><p>∫ (2tan(θ)sec²(θ)dθ)/((tan⁴(θ) + 1)·tan(θ)) = ∫ (2sec²(θ)dθ)/(tan⁴(θ) + 1)</p><p><strong>Step 3:</strong> Rewrite denominator: tan⁴(θ) + 1 = (tan²(θ))² + 1. Using tan²(θ) = x, this becomes unwieldy unless we convert strategically.</p><p><strong>Step 4:</strong> Alternative approach: Let u = √x, then x = u², dx = 2u·du</p><p>∫ (2u·du)/(u⁴ + 1)·u = ∫ (2du)/(u⁴ + 1)</p><p><strong>Step 5:</strong> For ∫ du/(u⁴ + 1), use the standard result: this integrates to (1/√2)arctan(u² - 1/u²)·(1/(2u)) after partial fractions or known formula</p><p><strong>Step 6:</strong> The standard antiderivative is: (1/(2√2))·arctan((x - 1/x)/√2) + C</p><p>Simplified: (1/(2√2))arctan(√x - 1/√x) + C</p><p>∴ Answer: <strong>(1/(2√2))arctan(√x - 1/(√x)) + C</strong> or equivalent form</p>
Correct Answer: A