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Mathematics
Mathematics
Problem
For
x
>
0
x>0
x
>
0
, determine whether
K
(
x
)
=
arctan
(
x
)
K(x)=\arctan(\sqrt{x})
K
(
x
)
=
arctan
(
x
)
is an antiderivative of
k
(
x
)
=
1
2
x
(
1
+
x
)
k(x)=\dfrac{1}{2\sqrt{x}\,(1+x)}
k
(
x
)
=
2
x
(
1
+
x
)
1
by finding
K
′
(
x
)
K'(x)
K
′
(
x
)
and comparing it with
k
(
x
)
k(x)
k
(
x
)
.
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