Evaluate

y = ∫(x2 + 2x + 1) ex dx

y = ∫(x2 + 2x)ex dx + ∫ex dx


We know that ∫(f(x) +f’(x))ex dx = f(x) ex


Here, f(x) = x2 then f’(x) = 2x


y = x2ex + ex + c


y = (x2 + 1)ex + c


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