Question

What is the value of ${\int }_0^a\frac{x-a}{x+a}dx$?

• A. $a+2a\log 2$
• B. $a-2a\log 2$
• C. $2a\log 2-a$
• D. $2a\log 2$

Answer 1

Answer: (B)

$a-2a\log 2$

$I{=}_0^a\int \frac{x-a}{x+a}dx$

Let $t=x+a\Rightarrow dt=dx$

When $x=0,t=a$

$x=a,t=2a$

Substituting, $I{=}_a^{2a}\int \frac{t-2a}{t}dt$

$=[t-2a\log |t|{]}_a^{2a}$

$=2a-2a\log 2a-a+2a\log a$

$=a-2a\log 2$

$\therefore$ Option $B$ is correct.