Question

If $\int\limits_{\frac{1}{3}}^{3}\left|\log _e x\right| d x=\frac{m}{n} \log e\left(\frac{n^2}{v}\right)$, where $m$ and $n$ are coprime natural numbers, then $m^2+n^2-5$ is equal to _____

Answer 1

$ \int\limits_{\frac{1}{3}}^3|\ell n x| d x=\int\limits_{\frac{1}{3}}^1(-\ell n x) d x+\int\limits_1^3(\ell n x) d x $
$=-[x \ell n x-x]_{1 / 3}^1+[x \ell n x-x]_1^3 $
$ =-\left[-1-\left(\frac{1}{3} \ell n \frac{1}{3}-\frac{1}{3}\right)\right]+[3 \ell n 3-3-(-1)] $
$ =\left[-\frac{2}{3}-\frac{1}{3} \ell n \frac{1}{3}\right]+[3 \ell n 3-2] $
$=-\frac{4}{3}+\frac{8}{3} \ell n 3$
$ =\frac{4}{3}(2 \ell n 3-1) $
$ =\frac{4}{3}\left(\ell n \frac{9}{ e }\right) $
$ \therefore m =4, n =3$
Now, $m ^2+ n ^2-5=16+9-5=20$