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Mathematics
Problem
The integral
∫
−
1
/
2
1
/
2
{
[
x
]
+
log
e
(
1
+
x
1
−
x
)
}
d
x
\int_{-1/2}^{1/2} \left\{ [x] + \log_e \left( \frac{1+x}{1-x} \right) \right\} dx
∫
−
1/2
1/2
{
[
x
]
+
lo
g
e
(
1
−
x
1
+
x
)
}
d
x
is equal to (where,
[
.
]
[. ]
[
.
]
denotes greatest integer function) a.
−
1
2
-\frac{1}{2}
−
2
1
b.
0
0
0
c.
1
1
1
d.
2
log
(
1
2
)
2 \log \left( \frac{1}{2} \right)
2
lo
g
(
2
1
)
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