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. $\log\left|\frac{x}{x+a}\right|+C$

Answer 1

Answer: (B)

$a\log\left|\frac{x+a}{x}\right|+C$

Answer 2

Let $I=\int\left(\frac{x-a}{x}-\frac{x}{x+a}\right) d x$

$=\int \frac{\left(x^{2}-a^{2}\right)-x^{2}}{x(x+a)} d x$

$=-a^{2} \int \frac{1}{x(x+a)} d x$

$=\frac{-a^{2}}{a} \int\left[\frac{1}{x}-\frac{1}{x+a}\right] d x$

$=-a \log \left|\frac{x}{x+a}\right|+C$

$=a \log \left|\frac{x+a}{x}\right|+C$

So, The correct option is B.