Question

$\int a^x[\log x+\log a\log \left(\frac{x^x}{e^x}\right)]dx$

• A. $x(\log x-1)$
• B. $a^x.x(\log x-1)$
• C. $a^x.x(\log x+1)$
• D. $a^x.(\log x-1)$

Answer 1

Answer: (B)

$a^x.x(\log x-1)$

$I=\int a^x\log xdx+\int \left(a^x\log a\right).x(\log x-1)dx$
We know that
$\int \log xdx=x\log x-\int x.\frac{1}{x}dx=x(\log x-1)$
Integrating 1st term by parts,
$\therefore I=a^x.x(\log x-1)-\int x(\log x-1)\left(a^x\log a\right)dx+\int (a^x.\log a).x(\log x-1)dx$
$=a^x.x(\log x-1)$