Typical Indefinite Integration 2

$\displaystyle \int \dfrac{\ln(x)^{\ln(x)}\left [\ln(x)\left (\ln(\ln(x))+1 \right )+1\right]}{x} \, \mathrm dx$

= $\displaystyle \int (\ln(x)^{\ln(x)}\left [\ln(x)\left (\ln(\ln(x))+1 \right )+1\right])\frac1x \, \mathrm dx$

Let $u=\ln(x)$ , then $\mathrm du=\frac1x\mathrm dx$ .

= $\displaystyle \int u^u[u(\ln(u)+1)+1]\,\mathrm du$

= $\displaystyle \int u^uu(\ln(u)+1)\,\mathrm du + \int u^u\,\mathrm du$

Let $y=u^u$ , then $\mathrm dy=u^u(\ln(u)+1)\mathrm du$ .

= $\displaystyle \int u \,\mathrm dy + \int y\,\mathrm du$

Integration by parts.

= $\displaystyle uy - \int y \,\mathrm du + \int y\,\mathrm du$

= $\displaystyle uy$

= $\displaystyle u\times u^u$

= $\displaystyle u^{u+1}$

= $\displaystyle \ln(x)^{\ln(x)+1}$

Therefore, $f(x) = \ln(x)$ and so, our answer turns out to be $\boxed{1}$