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Mathematics Question on Integration by Parts

The integral ∫−1/21/2([x]+log⁡(1+x1−x)) dx\int\limits^{1/2}_{-1/2} \bigg ([x] + \log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx equals

A

−12-\frac{1}{2}

B

0

C

1

D

log(12)log \bigg ( \frac{1}{2} \bigg )

Answer

−12-\frac{1}{2}

Explanation

Solution

∫−1/21/2([x]+log(1+x1−x)) dx \int ^{1/2}_{-1/2} \bigg ([x] + log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx \,
=∫−1/21/2[x] dx+∫−1/21/2 log(1+x1−x)dx= \int^{1/2}_{-1/2} [x] \, dx + \int^{1/2}_{-1/2} \, log \bigg (\frac{1+ x }{1 - x } \bigg ) dx
=∫−1/21/2[x] dx+0              [∵(1+x1−x)= \int^{1/2}_{-1/2} [x] \, dx + 0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bigg [\because \bigg ( \frac{ 1 + x }{ 1 - x} \bigg ) is an odd function ]\bigg ]
=∫−1/20[x] dx+∫01/2[x] dx=∫−1/20(−1) dx+∫01/2 dx= \int^{0}_{-1/2} [x] \, dx + \int^{1/2}_{0} [x] \, dx = \int^{0}_{-1/2} ( - 1) \, dx + \int^{1/2}_{0} \, dx
=[x]−1/20=−(0+12)=−12=[ x]^0_{-1/2} = - \bigg ( 0 + \frac{1}{2} \bigg ) =- \frac{1}{2}