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Question: The integral \(\int _ { - 1 / 2 } ^ { 1 / 2 } \left( [ x ] + \ln \left( \frac { 1 + x } { 1 - x } \...

The integral ∫−1/21/2([x]+ln⁡(1+x1−x))dx\int _ { - 1 / 2 } ^ { 1 / 2 } \left( [ x ] + \ln \left( \frac { 1 + x } { 1 - x } \right) \right) d x equal to

A

−12\frac { - 1 } { 2 }

B

0

C

1

D

2ln⁡(12)2 \ln \left( \frac { 1 } { 2 } \right)

Answer

−12\frac { - 1 } { 2 }

Explanation

Solution

log⁡(1+x1−x)\log \left( \frac { 1 + x } { 1 - x } \right) is an odd function of x as f(−x)=−f(x)f ( - x ) = - f ( x )

I=∫−1/21/2[x]dx+0I = \int _ { - 1 / 2 } ^ { 1 / 2 } [ x ] d x + 0 ⇒ I=∫−1/20[x]dx+∫01/2[x]dxI = \int _ { - 1 / 2 } ^ { 0 } [ x ] d x + \int _ { 0 } ^ { 1 / 2 } [ x ] d x

⇒ I=∫−1/20−1dx+0I = \int _ { - 1 / 2 } ^ { 0 } - 1 d x + 0 ⇒ −[x]−1/20=−12- [ x ] _ { - 1 / 2 } ^ { 0 } = \frac { - 1 } { 2 }.