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Q.
$\int\limits_{2-\log 3}^{3+\log 3} \frac{\log (4+x)}{\log (4+x)+\log (9-x)} d x$ is equal to :
Integrals
Solution:
$I=\int\limits_{2-\log 3}^{3+\log 3} \frac{\log (4+x)}{\log (4+x)+\log (9-x)} d x ....$(1)
using $ \int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x $
$ I=\int\limits_{2-\log 3}^{3+\log 3} \frac{\log (9-x)}{\log (9-x)+\log (4+x)} d x .....$(2)
From equation (1) and (2)
$ 2 I =\int\limits_{2-\log 3}^{3+\log 3} dx =1+2 \log 3 $
$\therefore I =\frac{1}{2}+\log 3$