Question

The integral $ \int\limits^{1/2}_{-1/2} \bigg ([x] + \log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx $ equals

• A. $-\frac{1}{2}$
• B. 0
• C. 1
• D. $log \bigg ( \frac{1}{2} \bigg )$

Answer 1

The Correct Option is A

$ \int ^{1/2}_{-1/2} \bigg ([x] + log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx \, $
$ = \int^{1/2}_{-1/2} [x] \, dx + \int^{1/2}_{-1/2} \, log \bigg (\frac{1+ x }{1 - x } \bigg ) dx $
$ = \int^{1/2}_{-1/2} [x] \, dx + 0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bigg [\because \bigg ( \frac{ 1 + x }{ 1 - x} \bigg ) $ is an odd function $\bigg ]$
$ = \int^{0}_{-1/2} [x] \, dx + \int^{1/2}_{0} [x] \, dx = \int^{0}_{-1/2} ( - 1) \, dx + \int^{1/2}_{0} \, dx$
$=[ x]^0_{-1/2} = - \bigg ( 0 + \frac{1}{2} \bigg ) =- \frac{1}{2}$