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Mathematics
Problem
If
∫
e
x
−
1
e
x
+
1
d
x
=
f
(
x
)
+
c
\int \frac{e^x - 1}{e^x + 1} dx = f(x) + c
∫
e
x
+
1
e
x
−
1
d
x
=
f
(
x
)
+
c
, then
f
(
x
)
f(x)
f
(
x
)
may be equal to (1)
2
log
∣
e
x
+
1
∣
2 \log |e^x + 1|
2
lo
g
∣
e
x
+
1∣
(2)
log
∣
e
2
x
−
1
∣
\log |e^{2x} - 1|
lo
g
∣
e
2
x
−
1∣
(3)
2
log
(
e
x
+
1
)
−
x
2 \log (e^x + 1) - x
2
lo
g
(
e
x
+
1
)
−
x
(4)
log
(
e
2
x
+
1
)
\log (e^{2x} + 1)
lo
g
(
e
2
x
+
1
)
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