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: \( \boxed{\log \left| \frac{x}{x + \cos x} \right| + C} \).

Answer 2

Step 1: Observe the structure of the integrand 

We are given: \[ I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} \, dx \] Notice that the denominator is: \[ x(x + \cos x) \] And the numerator: \[ \cos x + x \sin x \] This resembles the derivative of the denominator. Let’s check.

Step 2: Let the denominator be \( u \)

Let: \[ u = x + \cos x \Rightarrow \frac{du}{dx} = 1 - \sin x \quad \text{(Not useful)} \] But try: \[ f(x) = \log \left| \frac{x}{x + \cos x} \right| \Rightarrow \frac{d}{dx} \left( \log \left| \frac{x}{x + \cos x} \right| \right) = \frac{d}{dx} \left( \log |x| - \log |x + \cos x| \right) \] Derivatives: \[ \frac{d}{dx}(\log |x|) = \frac{1}{x}, \quad \frac{d}{dx}(\log |x + \cos x|) = \frac{1 - \sin x}{x + \cos x} \] So: \[ \frac{d}{dx} \left( \log \left| \frac{x}{x + \cos x} \right| \right) = \frac{1}{x} - \frac{1 - \sin x}{x + \cos x} \] Now make common denominator: \[ = \frac{(x + \cos x) - x(1 - \sin x)}{x(x + \cos x)} = \frac{x + \cos x - x + x \sin x}{x(x + \cos x)} = \frac{\cos x + x \sin x}{x(x + \cos x)} \] Which matches the given integrand.

Step 3: Final integration

Therefore, \[ \int \frac{\cos x + x \sin x}{x (x + \cos x)} \, dx = \log \left| \frac{x}{x + \cos x} \right| + C \]

Answer:

\( \boxed{ \log \left| \frac{x}{x + \cos x} \right| + C } \)