Question

$\int \left( \frac{x-a}{x} - \frac{x}{x+a} \right)\, dx$ is equal to:

• A. $\log \left| \frac{x+a}{x} \right| + C$
• B. $a \log \left| \frac{x+a}{x} \right| + C$
• C. $a \log \left| \frac{x}{x+a} \right| + C$
• D. $\log \left| \frac{x}{x+a} \right| + C$
• E. $a \log \left| \frac{x-a}{x+a} \right| + C$

Hint:

Algebraic simplification is often the most important step in integration. Reducing rational fractions into a constant and a simple denominator term makes the final integration instantaneous using logarithmic rules.

Answer 1

The Correct Option is B

Concept: To solve this integral, first simplify each term of the integrand using basic division or algebraic manipulation to make them easier to integrate using the $\int \frac{1}{u} du = \log|u| + C$ rule.

Step 1: Simplify the integrand.
Expand and simplify the terms:
\[ \frac{x-a}{x} = \frac{x}{x} - \frac{a}{x} = 1 - \frac{a}{x} \] For the second term, use the $+a, -a$ trick: \[ \frac{x}{x+a} = \frac{x + a - a}{x+a} = \frac{x+a}{x+a} - \frac{a}{x+a} = 1 - \frac{a}{x+a} \]

Step 2: Combine the simplified terms.
The full integrand becomes:
\[ \left( 1 - \frac{a}{x} \right) - \left( 1 - \frac{a}{x+a} \right) \] \[ = 1 - \frac{a}{x} - 1 + \frac{a}{x+a} \] \[ = \frac{a}{x+a} - \frac{a}{x} \]

Step 3: Perform the integration.
\[ \int \left( \frac{a}{x+a} - \frac{a}{x} \right) dx = a \log|x+a| - a \log|x| + C \] Using the property $\log M - \log N = \log \frac{M}{N}$: \[ = a \log \left| \frac{x+a}{x} \right| + C \]