Question

If $\underset{1}{\overset{3}{\int }}\frac{xdx}{2x^2+3[x{]}^2}=\frac{1}{4}(a\cdot \ln 2+b\ln 3+c\ln 5+d\ln 11),$ then which of the following statement is/are true $?$
(where $[\cdot ]$ denotes the greatest integer function)

• A. $a+b=0$
• B. $b+c=0$
• C. $c+d=0$
• D. $a+b+c=0$

Answer 1

Answer: (A), (B), (C)

$c+d=0$

Let $I=\underset{1}{\overset{3}{\int }}\frac{xdx}{2x^2+3[x{]}^2}=\underset{1}{\overset{2}{\int }}\frac{xdx}{2x^2+3\times 1}+\underset{2}{\overset{3}{\int }}\frac{xdx}{2x^2+3\times 4}$
$=\frac{1}{4}{\left(\ln |2x^2+3|\right)}_1^2+\frac{1}{4}{(\ln \mid 2x^2+12)}_2^3$

$=\frac{1}{4}(\ln 11-\ln 5+\ln 30-\ln 20)$

$=\frac{1}{4}(-\ln 2+\ln 3-\ln 5+\ln 11)$

$\Rightarrow a=-1,b=1,c=-1,d=1$

so, $a+b=0,b+c=0,c+d=0$