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Integrate  \int_{1}^{2}\frac{\ln x}{\ln x + \ln(3-x)}dx

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As we learnt 

 

Properties of Definite Integration -

 

\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx

\int_{a}^{b}\frac{f\left ( x \right )}{f\left ( x \right )+f\left ( a+b-x \right )}dx= \int_{a}^{b}f\left ( a+b-x \right )dx= \frac{b-a}{2}

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 Thus \: I=\frac{2-1}{2}= \frac{1}{2}

 


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