Question

${\int }_0^{\frac{a}{2}}\frac{adx}{(x-a)(x-2a)}$ equals to

• A. $\ln \frac{2}{3}$
• B. $\ln \frac{3}{2}$
• C. $\ln 6$
• D. $none$

Answer 1

Answer: (B)

$\ln \frac{3}{2}$

${\int }_0^{\frac{a}{2}}\frac{a}{\left(x-a\right)\left(x-2a\right)}dx$

$\frac{a}{\left(x-a\right)\left(x-2a\right)}=\frac{A}{x-a}+\frac{B}{x-2a}$

$\Rightarrow a=A\left(x-2a\right)+B\left(x-a\right)$

$\Rightarrow a=\left(A+B\right)x-2aA-aB$

on comparing both sides we get

$A+B=0\Rightarrow A=-B$ --(i)

and $-2aA-aB=a$

$\Rightarrow 2aB-aB=a$ (from (i))

$\Rightarrow B=1$

So, $A=-B=-1$

Now, ${\int }_0^{\frac{a}{2}}\frac{a}{\left(x-a\right)\left(x-2a\right)}dx$

$=-{\int }_0^{\frac{a}{2}}\frac{1}{x-a}dx+{\int }_0^{\frac{a}{2}}\frac{1}{x-2a}dx$

$={\left[-\ln \left|x-a\right|+\ln \left|x-2a\right|\right]}_0^{\frac{a}{2}}$

$={\left[\ln \left|\frac{x-2a}{x-a}\right|\right]}_0^{\frac{a}{2}}$

$=\ln \left|\frac{\frac{a}{2}-2a}{\frac{a}{2}-a}\right|-\ln \left|\frac{-2a}{-a}\right|$

$=\ln 3-\ln 2$

$=\ln \frac{3}{2}$