Question

$\underset{2-\log 3}{\overset{3+\log 3}{\int }}\frac{\log \left(4+x\right)}{\log \left(4+x\right)+\log \left(9-x\right)}dx$.

• A. Cannot be evaluated
• B. is equal to 5/2
• C. Is equal to 1+2 log 3
• D. is equal to 1/2+log 3

Answer 1

The correct option is D is equal to 1/2+log 3

$I={\int }_{2-log3}^{3+log3}\frac{log(4+x)}{log(4+x)+log(9-x)}dx$ ___ (1)

$I={\int }_{2-log3}^{3+log3}\frac{log(4+5-x)dx}{log(4+5-x)+log(9-5+x)}={\int }_{2-log3}^{3+log3}\frac{log(9-x)}{log(4+x)+log(9-x)}dx$ ___ (2)

$[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx]$

Adding (1), (2)

$\Rightarrow 2I={\int }_{2-log3}^{3+log3}\frac{log(x+4)+log(9-x)}{log(x+4)+log(9-x)}dx=[x{]}_{2-log3}^{3+log3}=3+log3-2+log3$

$=1+2log3$

$\Rightarrow I=\frac{1}{2}+log3$

$\therefore$ Option D is correct