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Problems
Let
∫
2
−
tan
x
3
+
tan
x
d
x
=
1
2
(
α
x
+
log
e
∣
β
sin
x
+
γ
cos
x
∣
)
+
C
\int \frac{2 - \tan x}{3 + \tan x}\,dx = \frac{1}{2}\,(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C
∫
3
+
t
a
n
x
2
−
t
a
n
x
d
x
=
2
1
(
α
x
+
lo
g
e
∣
β
sin
x
+
γ
cos
x
∣
)
+
C
…
Mathematics
Problem
Let
∫
2
−
tan
x
3
+
tan
x
d
x
=
1
2
(
α
x
+
log
e
∣
β
sin
x
+
γ
cos
x
∣
)
+
C
\int \frac{2 - \tan x}{3 + \tan x}\,dx = \frac{1}{2}\,(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C
∫
3
+
t
a
n
x
2
−
t
a
n
x
d
x
=
2
1
(
α
x
+
lo
g
e
∣
β
sin
x
+
γ
cos
x
∣
)
+
C
, where
C
C
C
is the constant of integration. Then,
α
+
γ
β
\alpha + \frac{\gamma}{\beta}
α
+
β
γ
is equal to (a) 1 (b) 3 (c) 7 (d) 4
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