int e^x(x+1)^2 dx= | vedclass

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Question

(A) We use the formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Let $f(x) = (x+1)^2 = x^2 + 2x + 1$. Then $f'(x) = 2x + 2 = 2(x+1)$. This does not

Vedclass · Accepted Answer

$e^x(x^2+1)+c$

Vedclass · Answer

(A) We use the formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Let $f(x) = (x+1)^2 = x^2 + 2x + 1$. Then $f'(x) = 2x + 2 = 2(x+1)$. This does not directly fit the form. Alternatively,expand the expression: $\int e^x(x^2+2x+1) dx = \int e^x x^2 dx + \int e^x(2x+1) dx$. Using integration by parts on $\int e^x x^2 dx$: $= x^2 e^x - \int 2x e^x dx + \int 2x e^x dx + \int e^x dx = x^2 e^x + e^x + c = e^x(x^2+1) + c$.

$\int e^x(x+1)^2 dx=$

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