Question

Evaluate the integral: \[ I = \int \frac{\cos x + x \sin x}{x (x + \cos x)} dx =\] 

• A. \( \log |x^2 + x \cos x| + C \)
• B. \( \log \left| \frac{x}{x + \cos x} \right| + C \)
• C. \( \log \left| \frac{\cos x}{x + \cos x} \right| + C \)
• D. \( \log \left| \frac{1}{x + \cos x} \right| - \log x + C \)

Hint:

For integrals involving expressions like \( x + \cos x \), consider using substitution where \( u = x + \cos x \) to simplify the terms.

Answer 1

The Correct Option is B

Step 1: Substituting \( u \)
Let: \[ u = x + \cos x. \] Differentiating both sides: \[ du = (1 - \sin x) dx. \] Rewriting the given integral: \[ I = \int \frac{\cos x + x \sin x}{x u} dx. \] Using the identity \( \cos x = \frac{d}{dx}(\sin x) \), we express: \[ \cos x + x \sin x = \frac{d}{dx} (x + \cos x). \] Step 2: Transforming the Integral
Rewriting the integral in terms of \( u \): \[ I = \int \frac{du}{x u}. \] Since \( du = (1 - \sin x) dx \), we substitute: \[ I = \int \frac{du}{x u} = \int \frac{dx}{x + \cos x}. \] Step 3: Evaluating the Integral
From standard logarithmic integration results, we obtain: \[ I = \log \left| \frac{x}{x + \cos x} \right| + C. \] Final Answer: \( \boxed{\log \left| \frac{x}{x + \cos x} \right| + C} \).