Question

$\underset{2-\log 3}{\overset{3+\log 3}{\int }}\frac{\log (4+x)}{\log (4+x)+\log (9-x)}dx$ is equal to :

• A. cannot be evaluated
• B. $\frac{5}{2}$
• C. $1+2\log 3$
• D. $\frac{1}{2}+\log 3$

Answer 1

Answer: (D)

$I=\underset{2-\log 3}{\overset{3+\log 3}{\int }}\frac{\log (4+x)}{\log (4+x)+\log (9-x)}dx....$(1)
using $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I=\underset{2-\log 3}{\overset{3+\log 3}{\int }}\frac{\log (9-x)}{\log (9-x)+\log (4+x)}dx.....$(2)
From equation (1) and (2)
$2I=\underset{2-\log 3}{\overset{3+\log 3}{\int }}dx=1+2\log 3$
$\therefore I=\frac{1}{2}+\log 3$