int_{2 - log 3}^{3 + log 3} frac{log (4 + x)} | vedclass

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Question

(A) Let $I = \int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we know that

Vedclass · Accepted Answer

$\frac{1}{2} + \log 3$

Vedclass · Answer

(A) Let $I = \int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we know that $I = \frac{b-a}{2}$. Here,$a = 2 - \log 3$ and $b = 3 + \log 3$. Also,$f(x) = \log(4+x)$,so $f(a+b-x) = \log(4 + (2 - \log 3 + 3 + \log 3 - x)) = \log(4 + 5 - x) = \log(9 - x)$. Thus,$I = \frac{(3 + \log 3) - (2 - \log 3)}{2} = \frac{1 + 2 \log 3}{2} = \frac{1}{2} + \log 3$.

$\int_{2 - \log 3}^{3 + \log 3} \frac{\log (4 + x)}{\log (4 + x) + \log (9 - x)} \, dx = $

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