Definite Integration Question 58
Question: $ \int _{-1}^{1}{\log (x+\sqrt{x^{2}+1})dx=} $
[MP PET 2001]
Options:
A) 0
B) log 2
C) $ \log \frac{1}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ f(x)=\log (x+\sqrt{1+x^{2}}) $
Now, $ f(-x)=\log ( \sqrt{1+x^{2}}-x )=\log (\sqrt{1+x^{2}}-x).\frac{(\sqrt{1+x^{2}}+x)}{(\sqrt{1+x^{2}}+x)} $
$ =\log \frac{[(1+x^{2})-x^{2}]}{(\sqrt{1+x^{2}}+x)} $
$ =\log 1-\log (\sqrt{1+x^{2}}+x) $
$ =-\log (\sqrt{1+x^{2}}+x) $
$ =-f(x) $
Hence, $ \int _{-1}^{1}{\log (x+\sqrt{1+x^{2}})=0} $ , $ [ \because \int _{-a}^{a}{f(x)=0,}\text{if }f(-x)=-f(x) ] $ .
BETA
