The correct option is B e^x1/x+c ∫ (x 1/x^2)e^x dx =∫ e^x(^1/_x 1/x)dx d(^1/x)/dx= ^1/x^2 so, It is in the form of ∫ e^x(f(x)+f^'(x))dx ...

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Integration by Partial Fractions

∫x-1/x2exdx=

Question

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

e^{x} + c

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e^{x \frac{1}{x}} + c

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- e^{x \frac{1}{x^{2}} + C}

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

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Solution

The correct option is B $e^{x \frac{1}{x}} + c$
$\int (\frac{x - 1}{x^{2}}) e^{x} d x$
$= \int e^{x} (^{1} /_{x} - 1 / x) d x$
$\frac{d (^{1} / x)}{d x} = -^{1} / x^{2}$
so, It is in the form of
$\int e^{x} (f (x) + f^{^{'}} (x)) d x = e^{x} f (x) + c$
$s o, \int e^{x} (1 / x - 1 / x^{2}) d x = e^{x} / x + c$

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Similar questions

\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x is equal to (a) \frac{e^{x}}{1 + x} + C (b) - \frac{e^{x}}{1 + x^{2}} + C (c) \frac{e^{x}}{(1 + x}

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