∫12x(x2−1)8dx
Let substitute, u=x2−1
Hence, dxdu=2x, dx=2x1du
=21∫u8du
Now solving, ∫u8du
Apply power rule
∫u8du=n+1un+1 With n=8
=18u9
Undo the substitution. u=x2−1
=18(x2−1)9
Replacing x with 2 and 1
=18(22−1)9−0=1093.5
False because 1093.5 is not a whole valued number
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