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Mathematics
Problem
∫
e
2
x
−
1
e
2
x
+
1
d
x
=
\int \frac{e^{2x}-1}{e^{2x}+1}\,dx =
∫
e
2
x
+
1
e
2
x
−
1
d
x
=
(a)
e
2
x
−
1
e
2
x
+
1
+
c
\frac{e^{2x}-1}{e^{2x}+1} + c
e
2
x
+
1
e
2
x
−
1
+
c
(b)
log
(
e
2
x
+
1
)
−
x
+
c
\log(e^{2x}+1) - x + c
lo
g
(
e
2
x
+
1
)
−
x
+
c
(c)
log
(
e
2
x
+
1
)
+
c
\log(e^{2x}+1) + c
lo
g
(
e
2
x
+
1
)
+
c
(d) None of these (a)
e
2
x
−
1
e
2
x
+
1
+
c
\frac{e^{2x}-1}{e^{2x}+1} + c
e
2
x
+
1
e
2
x
−
1
+
c
(b)
log
(
e
2
x
+
1
)
−
x
+
c
\log(e^{2x}+1) - x + c
lo
g
(
e
2
x
+
1
)
−
x
+
c
(c)
log
(
e
2
x
+
1
)
+
c
\log(e^{2x}+1) + c
lo
g
(
e
2
x
+
1
)
+
c
(d) None of these
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