JEE MainMathematicsModerateDefinite Integration

$\int_{2 - log3}^{3 + log3}\mspace{2mu}\frac{log(4 + x)}{log(4 + x) + log(9 - x)}$
dx is equal to

Show answer and solution

I = \int_{2 - log3}^{3 + log3}\mspace{2mu}\frac{log(4 + x)}{log(4 + x) + log(9 - x)}dx

.......(1)
Using

\int_{a}^{b}\mspace{2mu} f(x)dx = \int_{a}^{b}\mspace{2mu} f(a + b - x)dx

I = \int_{2 - log3}^{3 + log3}\mspace{2mu}\frac{log(9 - x)}{log(9 - x) + log(4 + x)}dx

........(2)
From equation (1) and (2)

2I = \int_{2 - \log 3}^{3 + \log 3}\mspace{2mu} dx = 1 + 2\log 3

\therefore

I =

\frac{1}{2}

+ log 3

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.