The correct option is B e^x(x 1)+c ∫ x e^x dx=x∫ e^x dx ∫ (e^x dx)dx =xe^x e^x+c =e^x(x 1)+c ∫ xe^xdx= e^x(x 1)+c ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

∫ x.exdx=

Question

$\int x . e^{x} d x =$

x . e^{x} + c

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

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

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

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Solution

The correct option is B $e^{x} (x - 1) + c$
$\int x e^{x} d x = x \int e^{x} d x - \int (e^{x} d x) d x$
$= x e^{x} - e^{x} + c$
$= e^{x} (x - 1) + c$
$\int x e^{x} d x = e^{x} (x - 1) + c$

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Q. Evaluate

\int x e^{x} d x

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∫x.exdx=

The correct option is B ex(x−1)+c∫x ex dx=x∫ex dx−∫ (ex dx)dx=xex−ex+c=ex(x−1)+c∫xexdx= ex(x−1)+c

$\int x . e^{x} d x =$

The correct option is B $e^{x} (x - 1) + c$
$\int x e^{x} d x = x \int e^{x} d x - \int (e^{x} d x) d x$
$= x e^{x} - e^{x} + c$
$= e^{x} (x - 1) + c$
$\int x e^{x} d x = e^{x} (x - 1) + c$