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Mathematics
Problem
If
I
=
∫
e
−
x
log
(
e
x
+
1
)
d
x
I = \int e^{-x}\log(e^x + 1)\,dx
I
=
∫
e
−
x
lo
g
(
e
x
+
1
)
d
x
, then
I
I
I
equals (a)
x
+
(
e
−
x
+
1
)
log
(
e
x
+
1
)
+
C
x + (e^{-x} + 1)\log(e^x + 1) + C
x
+
(
e
−
x
+
1
)
lo
g
(
e
x
+
1
)
+
C
(b)
x
+
(
e
x
+
1
)
log
(
e
x
+
1
)
+
C
x + (e^x + 1)\log(e^x + 1) + C
x
+
(
e
x
+
1
)
lo
g
(
e
x
+
1
)
+
C
(c)
x
−
(
e
−
x
+
1
)
log
(
e
x
+
1
)
+
C
x - (e^{-x} + 1)\log(e^x + 1) + C
x
−
(
e
−
x
+
1
)
lo
g
(
e
x
+
1
)
+
C
(d) none of these (a)
x
+
(
e
−
x
+
1
)
log
(
e
x
+
1
)
+
C
x + (e^{-x} + 1)\log(e^x + 1) + C
x
+
(
e
−
x
+
1
)
lo
g
(
e
x
+
1
)
+
C
(b)
x
+
(
e
x
+
1
)
log
(
e
x
+
1
)
+
C
x + (e^x + 1)\log(e^x + 1) + C
x
+
(
e
x
+
1
)
lo
g
(
e
x
+
1
)
+
C
(c)
x
−
(
e
−
x
+
1
)
log
(
e
x
+
1
)
+
C
x - (e^{-x} + 1)\log(e^x + 1) + C
x
−
(
e
−
x
+
1
)
lo
g
(
e
x
+
1
)
+
C
(d) none of these
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