### $f(x) = {\left( {\frac{x}{\pi }} \right)^x} + {\left( {\frac{\pi }{x}} \right)^x}$ $\int {f(x)dx + } \int {f(x)\ln xdx} - \int {f(x)\ln \pi dx} = ?$

Combining given three integrals yields $\int {f(x)[1 + \ln x - \ln \pi ]dx}$

$= \int {f(x)(\ln e + \ln x - \ln \pi )dx}$

$= \int {\left[ {{{\left( {\frac{x}{\pi }} \right)}^x} + {{\left( {\frac{\pi }{x}} \right)}^x}} \right]\left( {\ln \frac{{ex}}{\pi }} \right)dx} = I$

Let, ${\left( {\frac{x}{\pi }} \right)^x} = t$

$\therefore x\ln \frac{x}{\pi } = \ln t$

$\therefore x(\ln x - \ln \pi ) = \ln t$

Differentiation yields $x.\frac{1}{x} + \ln x - \ln \pi = \frac{1}{t}.\frac{{dt}}{{dx}}$

$\therefore (1 + \ln x - \ln \pi )dx = \frac{{dt}}{t}$

$\therefore (\ln e + \ln x - \ln \pi )dx = \frac{{dt}}{t}$

$\Rightarrow \ln \left( {\frac{{ex}}{\pi }} \right)dx = \frac{{dt}}{t}$

$I = \int {\left( {t + \frac{1}{t}} \right)\frac{{dt}}{t}}$

$= \int {\left( {1 + \frac{1}{{{t^2}}}} \right)dt}$

$= t - \frac{1}{t} + C$

$= {\left( {\frac{x}{\pi }} \right)^x} - {\left( {\frac{\pi }{x}} \right)^x} + C$