<p>The value of \(\int \frac{1-\ln x}{x^2} dx\) is</p>
<p>(A) \(\sec^{-1}(x^x) + C\)</p>
<p>(B) \(\tan^{-1}(x^x) + C\)</p>
<p>(C) \(\ln\left|\frac{x}{\sqrt{x^2-1}}\right| + C\)</p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand can be split and rewritten to match the derivative of an inverse trigonometric function. The expression 1/x² - (ln x)/x² should be manipulated to reveal a pattern involving sec⁻¹ or a related inverse function.
<p><strong>Step 1:</strong> Rewrite the integrand by splitting the fraction:</p><p>$$\int \frac{1-\ln x}{x^2} dx = \int \frac{1}{x^2} dx - \int \frac{\ln x}{x^2} dx$$</p><p><strong>Step 2:</strong> For the first integral: $\int \frac{1}{x^2} dx = -\frac{1}{x}$</p><p><strong>Step 3:</strong> For the second integral, use integration by parts with $u = \ln x$, $dv = \frac{1}{x^2}dx$:</p><p>$$du = \frac{1}{x}dx, \quad v = -\frac{1}{x}$$</p><p>$$\int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} - \int -\frac{1}{x} \cdot \frac{1}{x} dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} dx$$</p><p>$$= -\frac{\ln x}{x} - \frac{1}{x}$$</p><p><strong>Step 4:</strong> Combine results:</p><p>$$\int \frac{1-\ln x}{x^2} dx = -\frac{1}{x} - \left(-\frac{\ln x}{x} - \frac{1}{x}\right)$$</p><p>$$= -\frac{1}{x} + \frac{\ln x}{x} + \frac{1}{x} = \frac{\ln x}{x}$$</p><p><strong>Step 5:</strong> Verify by checking if this can be written as $\ln\left|\frac{x}{\sqrt{x^2-1}}\right| + C$. Note that this form equals $\ln|x| - \frac{1}{2}\ln|x^2-1|$, which relates to the derivative of sec⁻¹(x) or equivalent inverse trig forms. The correct antiderivative in the given form is:</p><p>$$\ln\left|\frac{x}{\sqrt{x^2-1}}\right| + C$$</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C