∫ax(lnx+lna. ln(xe)x)dx= (where ‘c’ is a constant of integration)

∫[axlnx+axln⏟ii. x(lnx−1)⏟i] dx =∫ax. lnx dx+[x(lnx−1)ax−∫[x.1x+(lnx−1)]ax] dx ==∫ax. lnx dx+[x(lnx/e)] ax−∫(lnx)axdx

∫ax(lnx+lna. ln(xe)x)dx= (where ‘c’ is a constant of integration)

∫[axlnx+axln⏟ii. x(lnx−1)⏟i] dx =∫ax. lnx dx+[x(lnx−1)ax−∫[x.1x+(lnx−1)]ax] dx ==∫ax. lnx dx+[x(lnx/e)] ax−∫(lnx)axdx

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$\int a^{x} (\ln x + \ln a . \ln {(\frac{x}{e})}^{x}) d x =$ (where ‘c’ is a constant of integration)

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$a^{x} \ln {(\frac{e}{x})}^{2 x} + C$

$a^{x} \ln {(\frac{x}{e})}^{x} + C$

$a^{x} + \ln {(\frac{x}{e})}^{x} + C$

$a^{x} \ln {(\frac{e}{x})}^{x} + C$

answer is B.

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Detailed Solution

$\int [a^{x} \ln x + \underset{i i}{\underset{⏟}{a^{x} \ln}} . \underset{i}{\underset{⏟}{x (\ln x - 1)}}] d x = \int a^{x} . \ln x d x + [x (\ln x - 1) a^{x} - \int [x . \frac{1}{x} + (\ln x - 1)] a^{x}] d x = = \int a^{x} . \ln x d x + [x (\ln x / e)] a^{x} - \int (\ln x) a^{x} d x$