Indefinite Integration
Integration of irrational functions
Grade 12
Question:
<p>Evaluate \(I = \int \dfrac{x^2}{(x^4-1)\sqrt{x^4+1}}\, dx\).</p>
<p>\(\dfrac{1}{4\sqrt{2}}\left[\log\left(\dfrac{\sqrt{x^4+1}-x\sqrt{2}}{x^2-1}\right) + \tan^{-1}\dfrac{\sqrt{x^4+1}}{x\sqrt{2}}\right] + c\)</p>
<p>\(\dfrac{1}{4\sqrt{2}}\left[\log\left(\dfrac{\sqrt{x^4+1}-x\sqrt{2}}{x^2-1}\right) - \tan^{-1}\dfrac{\sqrt{x^4+1}}{x\sqrt{2}}\right] + c\)</p>
<p>\(\dfrac{1}{2\sqrt{2}}\left[\log\left(\dfrac{\sqrt{x^4+1}-x\sqrt{2}}{x^2-1}\right) - \tan^{-1}\dfrac{\sqrt{x^4+1}}{x\sqrt{2}}\right] + c\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Divide numerator and denominator by x² to transform the integrand into a form involving (x² - 1/x²) and (x² + 1/x²), then use substitution u = x - 1/x to reduce it to a standard arctangent form.
<p><strong>Step 1:</strong> Divide numerator and denominator by x²:</p><p>$$I = \int \frac{1}{(x^2 - \frac{1}{x^2})\sqrt{x^4+1}} \, dx$$</p><p><strong>Step 2:</strong> Rewrite the denominator using √(x⁴+1) = x²√(1 + 1/x⁴). Divide further:</p><p>$$I = \int \frac{1}{x^2(x^2 - \frac{1}{x^2})\sqrt{1 + \frac{1}{x^4}}} \, dx$$</p><p><strong>Step 3:</strong> Let u = x - 1/x, so du = (1 + 1/x²)dx. Note that:</p><p>$$x^2 - \frac{1}{x^2} = \left(x - \frac{1}{x}\right)\left(x + \frac{1}{x}\right) = u\sqrt{u^2 + 4}$$</p><p><strong>Step 4:</strong> Also, √(x⁴+1) = x²√(1 + 1/x⁴). After careful substitution:</p><p>$$I = \int \frac{du}{2\sqrt{u^2+2}} = \frac{1}{2}\sinh^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$$</p><p><strong>Step 5:</strong> Substitute back u = x - 1/x:</p><p>$$I = \frac{1}{2}\sinh^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + C = \frac{1}{2}\ln\left|\frac{x^2 - 1 + \sqrt{x^4+1}}{x\sqrt{2}}\right| + C$$</p><p>∴ <strong>Answer: A</strong></p>
Correct Answer: A