Question

∫ eˣ(1+x log x/x) dx=.......+c ,Select correct option from the given options.
(a) eˣ log x
(b) x.eˣ
(c) 1/x log x
(d) e⁻ˣ log x

Answer 1

we have to find out the value of $\int{e^x\{\frac{1+xlogx}{x}\}}\,dx$

$\int{e^x\{\frac{1}{x}+\frac{xlogx}{x}\}}\,dx$

= $\int{e^x\{\frac{1}{x}+logx\}}\,dx$

now, if we let logx = f(x)
then, 1/x = f'(x) [ because differentiate of logx is 1/x ]

so, it seems like $\int{e^x\{f'(x)+f(x)\}}\,dx$

we know, $\int{e^{(ax+b)}\{f(x)+f'(x)\}}\,dx=\frac{1}{a}e^{(ax+b)}f(x)+C$

so, $\int{e^x\{\frac{1}{x}+logx\}}\,dx=e^xlogx+C$

hence, option (a) is correct.