If ∫ 1 3 3 log e x dx = m n log e n 2 e , where m and n are coprime natural numbers, then m 2 + n 2 -…

Let

$I={\int }_{\frac{1}{3}}^3\left|\ln x\right|dx$

$\Rightarrow I={\int }_{\frac{1}{3}}^1\left|\ln x\right|dx+{\int }_1^3\left|\ln x\right|dx$

$\Rightarrow I=-{\int }_{\frac{1}{3}}^1\ln xdx+{\int }_1^3\ln xdx$

Since, $\ln x<0 \mathrm{for} x\in \left(0,1\right)$.

$\Rightarrow I=-{\left[x\ln x-x\right]}_{\frac{1}{3}}^1+{\left[x\ln x-x\right]}_1^3$

$\Rightarrow I=-\left[\left(0-1\right)-\left(-\frac{1}{3}\ln 3-\frac{1}{3}\right)\right]+\left[3\ln 3-3-\left(0-1\right)\right]$

$\Rightarrow I=\frac{2}{3}-\frac{1}{3}\ln 3+3\ln 3-2$

$\Rightarrow I=\frac{8}{3}\ln 3-\frac{4}{3}$

$\Rightarrow I=\frac{4}{3}\left(2\ln 3-1\right)$

$\Rightarrow I=\frac{4}{3}\left(\ln 3^2-\ln e\right)$

$\Rightarrow I=\frac{4}{3}\ln \left(\frac{3^2}{e}\right)$

So, $m=4, n=3$

Hence, $m^2+n^2-5=16+9-5=20$