The correct option is D e^x(x^2 4x+5)+c ∫ e^x(x 1)^2dx ∫ e^x(x^2 2x+1)dx x^2 2x+1=(ax^2+bx+c)+(2ax+b) a=1; b+2= 2; b= 4, c+b=1, c=5 ...

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Integration by Parts

∫ exx-12dx=

Question

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

e^{x} (x - 1) + c

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e^{x} (x^{2} - 4 x) + c

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e^{x} (x^{2} + 4 x - 5) + c

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e^{x} (x^{2} - 4 x + 5) + c

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Solution

The correct option is D $e^{x} (x^{2} - 4 x + 5) + c$
$\int e^{x} (x - 1)^{2} d x$
$\int e^{x} (x^{2} - 2 x + 1) d x$
$x^{2} - 2 x + 1 = (a x^{2} + b x + c) + (2 a x + b)$
$a = 1; b + 2 = - 2; b = - 4, c + b = 1, c = 5$
$\int e^{x} (x^{2} - 2 x + 1) d x = \int e^{x} [(x^{2} - 4 x + 5) + (2 x - 4)] d x$
It is in the form of
$\int e^{x} [f (x) + f^{1} (x)] d x = e^{x} f (x) + c$
$= e^{x} (x^{2} - 4 x + 5) + c$

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