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Mathematics
Problem
If
∫
2
e
x
+
3
e
−
x
3
e
x
+
4
e
−
x
d
x
=
A
x
+
B
log
(
3
e
2
x
+
4
)
+
C
\int \frac{2e^x + 3e^{-x}}{3e^x + 4e^{-x}}\,dx = Ax + B \log(3e^{2x} + 4) + C
∫
3
e
x
+
4
e
−
x
2
e
x
+
3
e
−
x
d
x
=
A
x
+
B
lo
g
(
3
e
2
x
+
4
)
+
C
, then (A)
A
=
−
3
4
,
B
=
1
24
A = -\frac{3}{4}, B = \frac{1}{24}
A
=
−
4
3
,
B
=
24
1
(B)
A
=
3
4
,
B
=
−
1
24
A = \frac{3}{4}, B = -\frac{1}{24}
A
=
4
3
,
B
=
−
24
1
(C)
A
=
1
4
,
B
=
1
24
A = \frac{1}{4}, B = \frac{1}{24}
A
=
4
1
,
B
=
24
1
(D)
A
=
−
3
4
,
B
=
1
4
A = -\frac{3}{4}, B = \frac{1}{4}
A
=
−
4
3
,
B
=
4
1
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