∫
x
n
d
x
=
x
n
+
1
n
+
1
\displaystyle \int{x}^{{n}}\ {\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}
∫
x
n
d
x
=
n
+
1
x
n
+
1
for any
n
\displaystyle {n}
n
.
False
True
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