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Mathematics
Problem
If
∫
4
+
x
2
x
6
d
x
=
A
(
4
+
x
2
)
3
/
2
(
B
x
2
−
6
)
x
5
+
C
\int \frac{\sqrt{4+x^2}}{x^6} dx = \frac{A(4+x^2)^{3/2}(Bx^2-6)}{x^5} + C
∫
x
6
4
+
x
2
d
x
=
x
5
A
(
4
+
x
2
)
3/2
(
B
x
2
−
6
)
+
C
, then a.
A
=
1
120
A = \frac{1}{120}
A
=
120
1
b.
B
=
1
B = 1
B
=
1
c.
A
=
−
1
120
A = -\frac{1}{120}
A
=
−
120
1
d.
B
=
−
1
B = -1
B
=
−
1
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