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Question:

The integral ∫42 logx/(2logx+log(36x+x²)) dx is equal to

0

2

3

1

Solution:

I=∫42(2logx)/(2logx+2log(6-x))dx
I=∫42logx/(logx+log(6-x))dx
Using the property ∫baf(x) dx=∫baf(a+b-x) dx we get,
I=∫42log(6-x)/(logx+log(6-x))dx
Adding the two integrals we get,
2I=∫421dx
Hence,2I=2
Hence,I=1