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Factorization

∫ x+12 e x d ...

Question

$\int {(x + 1)}^{2} e^{x} d x$

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Solution

$\int \underset{I}{{(x + 1)}^{2}} \underset{II}{e^{x}} d x = {(x + 1)}^{2} \int e^{x} d x - \int \{\frac{d}{d x} {(x + 1)}^{2} \int e^{x} d x\} d x = {(x + 1)}^{2} \cdot e^{x} - \int 2 (x + 1) \cdot e^{x} d x = {(x + 1)}^{2} e^{x} - 2 \int \underset{I}{x} \underset{II}{e^{x}} d x - 2 \int e^{x} d x = {(x + 1)}^{2} e^{x} - 2 [x \cdot e^{x} - \int 1 \cdot e^{x} d x] - 2 e^{x} = {(x + 1)}^{2} e^{x} - 2 x e^{x} + 2 e^{x} - 2 e^{x} + C = [{(x + 1)}^{2} - 2 x] e^{x} + C = (x^{2} + 1) e^{x} + C$

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