∫ ( e x ( x -1)( x - ln x )/ x 2) dx is equal to (where c is constant of integration)

Question

∫ ( e x ( x -1)( x - ln x )/ x 2) dx is equal to (where c is constant of integration) (A) e x (( x - ln x / x ))+ c (B) e x (( x - ln x +1/ x ))+ c (C) e x (( x - ln x / x 2))+ c (D) e x (( x - ln x -1/ x ))+ c

Tardigrade Admin · Accepted Answer

∫ ( e x ( x -1)( x - ln x )/ x 2) dx =∫ e x ((1/ x )-(1/ x 2)) ln (( e x / x )) dx Put t=(ex/x) =∫ ln t dt =( t ln t - t )+ c =( e x / x )( ln (( e x / x ))-1)+ c =( e x ( x - ln x -1)/ x )+ c

Q. $\int \frac{e ^{x} ( x - 1 ) ( x - l n x )}{x ^{2}} d x$ is equal to
(where $c$ is constant of integration)

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

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

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

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

$\int \frac{e ^{x} ( x - 1 ) ( x - l n x )}{x ^{2}} d x = \int e^{x} (\frac{1}{x} - \frac{1}{x ^{2}}) ln (\frac{e ^{x}}{x}) d x$
Put $t = \frac{e ^{x}}{x}$
$= \int ln t d t = (t ln t - t) + c = \frac{e ^{x}}{x} (ln (\frac{e ^{x}}{x}) - 1) + c$
$= \frac{e ^{x} ( x - l n x - 1 )}{x} + c$

Q. ∫x2ex(x−1)(x−lnx)​dx is equal to (where c is constant of integration)

ex(xx−lnx​)+c

ex(xx−lnx+1​)+c

ex(x2x−lnx​)+c

ex(xx−lnx−1​)+c