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Problems
If
∫
(
x
2010
+
x
804
+
x
402
)
(
2
x
1608
+
5
x
402
+
10
)
1
/
402
d
x
=
1
10
a
(
2
x
2010
+
5
x
804
+
10
x
402
)
a
/
402
+
C
\int (x^{2010} + x^{804} + x^{402})(2x^{1608} + 5x^{402} + 10)^{1/402} dx = \frac{1}{10a} (2x^{2010} + 5x^{804} + 10x^{402})^{a/402} + C
∫
(
x
2010
+
x
804
+
x
402
)
(
2
x
1608
+
5
x
402
+
10
)
1/402
d
x
=
10
a
1
(
2
x
2010
+
5
x
804
+
10
x
402
)
a
/402
+
C
…
Mathematics
Problem
If
∫
(
x
2010
+
x
804
+
x
402
)
(
2
x
1608
+
5
x
402
+
10
)
1
/
402
d
x
=
1
10
a
(
2
x
2010
+
5
x
804
+
10
x
402
)
a
/
402
+
C
\int (x^{2010} + x^{804} + x^{402})(2x^{1608} + 5x^{402} + 10)^{1/402} dx = \frac{1}{10a} (2x^{2010} + 5x^{804} + 10x^{402})^{a/402} + C
∫
(
x
2010
+
x
804
+
x
402
)
(
2
x
1608
+
5
x
402
+
10
)
1/402
d
x
=
10
a
1
(
2
x
2010
+
5
x
804
+
10
x
402
)
a
/402
+
C
. Then
(
a
−
400
)
(a - 400)
(
a
−
400
)
is equal to .....
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