- e x x -1 x -ln x - x 2 dx is equal toA- e x x -ln x - x - cB- e x x -In x - x 2- cC- e x x -ln x-1- x - cD- e x x -ln x -1- x - c

The correct option is D e^x (x ln x 1/x) + c ∫e^x (x 1)(x ln x)/x^2 dx = ∫ e^x (1/x 1/x^2) (ln e^x/x)dx = ∫ ln e^x/x d (e^x/x) = ln (e^x/x) e^x/x e^x/x + c ...

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Byju's Answer

Standard XII

Physics

Definite Integrals

∫ e x x -1 x ...

Question

$\int \frac{e^{x} (x - 1) (x - l n x)}{x^{2}} d x$ is equal to

$e^{x} (\frac{x - l n x}{x}) + c$

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$e^{x} (\frac{x - l n x + 1}{x}) + c$

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$e^{x} (\frac{x - l n x}{x^{2}}) + c$

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$e^{x} (\frac{x - l n x - 1}{x}) + c$

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Solution

The correct option is D
$e^{x} (\frac{x - l n x - 1}{x}) + c$

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

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

$= \int l n \frac{e^{x}}{x} d (\frac{e^{x}}{x})$

$= l n (\frac{e^{x}}{x}) \frac{e^{x}}{x} - \frac{e^{x}}{x} + c$

$= \frac{e^{x}}{x} (x - l n x - 1) + c$

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∫ex(x−1)(x−ln x)x2dx is equal to

The correct option is D ex(x−ln x−1x)+c ∫ex(x−1)(x−ln x)x2dx =∫ex(1x−1x2)(lnexx)dx =∫lnexxd(exx) =ln(exx)exx−exx+c =exx(x−ln x−1)+c

$\int \frac{e^{x} (x - 1) (x - l n x)}{x^{2}} d x$ is equal to