calculus I integral
posted by John .
Is this correct?
Evaluate 2x∫1 3t(t^2 + 1)^2 dt
u= t^2 + 1
du= 2t dt 1/2du=tdt
2x∫1 3u^2 1/2du
3/2 2x∫1 1/3u^3 + c
3/2 [1/3 u^3]
3/2(1/3(2x)^3)  3/2(1/3(1)^3)
Answer= 4x  1/2
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