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If
∫
4
+
x
12
(
x
6
−
2
x
3
+
2
)
d
x
=
x
α
α
+
x
β
γ
+
δ
x
+
c
\int \frac{4 + x^{12}}{(x^6 - 2x^3 + 2)}\,dx = \frac{x^\alpha}{\alpha} + \frac{x^\beta}{\gamma} + \delta x + c
∫
(
x
6
−
2
x
3
+
2
)
4
+
x
12
d
x
=
α
x
α
+
γ
x
β
+
δ
x
+
c
…
Mathematics
Problem
If
∫
4
+
x
12
(
x
6
−
2
x
3
+
2
)
d
x
=
x
α
α
+
x
β
γ
+
δ
x
+
c
\int \frac{4 + x^{12}}{(x^6 - 2x^3 + 2)}\,dx = \frac{x^\alpha}{\alpha} + \frac{x^\beta}{\gamma} + \delta x + c
∫
(
x
6
−
2
x
3
+
2
)
4
+
x
12
d
x
=
α
x
α
+
γ
x
β
+
δ
x
+
c
, then the last digit of the number
(
α
−
β
+
γ
−
δ
)
99
(\alpha - \beta + \gamma - \delta)^{99}
(
α
−
β
+
γ
−
δ
)
99
is equal to (where 'c' is integration constant)
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