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∫ x eˣ dx (using integration by parts):

Ax² eˣ/2 + C
Bx² eˣ + C
Cx eˣ + C
D(x - 1) eˣ + C
Answer & Solution
Correct answer: D. (x - 1) eˣ + C
**By parts:** u = x, dv = eˣ dx, du = dx, v = eˣ. We have ∫ u dv = uv - ∫ v. Then du = x eˣ - ∫ eˣ. Then dx = x eˣ - eˣ + C. So dx = (x-1) eˣ + C.
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