Integrating a Logarithmic Function: How to Solve a Tricky Definite Integral

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Hysteria X
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Homework Statement


Evaluate:

##\int -1 to 1 log((2-x)/(2+x)) dx##

Homework Equations



Equation chain: integral from a to b of f(x) dx equals integral from a to b of d/dx g(x) dx, equals g(x) evaluated from a to b, equals g(b) minus g(a).

The Attempt at a Solution



okay so normal attempt at integration won't work coz then we get log 1
what am i supposed to do i tried applying some general formulas of definite integral but i don't seem to get the answer :confused:
 
Hysteria X said:

Homework Statement


Evaluate:

##\int -1 to 1 log((2-x)/(2+x)) dx##



Homework Equations



Equation chain: integral from a to b of f(x) dx equals integral from a to b of d/dx g(x) dx, which equals g(b) minus g(a), with g(x) evaluated from a to b.


The Attempt at a Solution



okay so normal attempt at integration won't work coz then we get log 1
what am i supposed to do i tried applying some general formulas of definite integral but i don't seem to get the answer :confused:

Can you show what you did using normal integration? I actually don't think log(1) is really wrong. But how did you get it?
 
This logarithmic identity should get you going:

## log\frac{a}{b} = log(a)-log(b) ##

I agree with Dick though, you have the correct answer.