Definite Integration Question 86
Question: The value of $ \int _{-2}^{2}{[ p\ln ( \frac{1+x}{1-x} )+q\ln {{( \frac{1-x}{1+x} )}^{-2}}+r ]dx} $ depends on
[Orissa JEE 2003]
Options:
A) The value of p
B) The value of q
C) The value of r
D) The value of p and q
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ \log ( \frac{1+x}{1-x} ) $ is an odd function
$ \therefore \int _{-2}^{2}{{ p\log ( \frac{1+x}{1-x} )+q\log {{( \frac{1-x}{1+x} )}^{-2}}+r }dx} $
$ =r\int _{-2}^{2}{dx}=4r. $
Hence depends on the value of r.
BETA
