Can you help me with the $x^{\ln(x)-1}\ln(x)$?

Question

Recently, I tried to solve $\displaystyle\int x^{\ln(x)-1}\ln(x)\,\mathrm dx$ then I was stuck on this part (but not on the substitution).

My stuck was when I saw in the calculator, that it does this:

$$\displaystyle\int x^{\ln(x)-1}\ln(x)\,\mathrm dx$$ Prepare for Substitution: $$\int\frac{1}{x}\cdot e^{\ln^2(x)}\ln(x)\,\mathrm dx$$

My question is, how $x^{\ln(x)-1}\ln(x)=\dfrac{1}{x}\cdot e^{\ln^2(x)}\ln(x)=\dfrac{e^{\ln^2(x)}\ln(x)}{x}$?

Answer 1

$$ \begin{aligned} \because x^{\ln x-1} = & \frac{x^{\ln x}}{x} \\ = & \frac{e^{\ln \left(x^{\ln x}\right)}}{x} \\ = & \frac{e^{\ln x(\ln x)}}{x} \\ = & \frac{e^{\ln ^2 x}}{x}\\ \therefore x^{\ln x-1} \ln x=&\frac{e^{\ln ^2 x} \ln x}{x} \end{aligned} $$

Answer 2

We know that (for $x>0$) $$ x = e^{\ln(x)} $$ so $$\begin{align*} x^{\ln(x)-1} &= e^{\ln ( x^{\ln(x)-1})}\\ &= e^{(\ln(x) - 1) \ln(x)}\\ &= e^{\ln^2(x) - \ln(x)}\\ &= e^{\ln^2(x)} e^{-\ln(x)}\\ &= \frac{1}{e^{\ln(x)}} e^{\ln^2(x)} \\ &= \frac 1 x e^{\ln^2(x)} \end{align*}$$

• The second equality uses the rule $\ln(x^r) = r \ln(x)$
• The third equality is just distribution
• The fourth equality uses the rule $a^{b+c} = a^b \cdot a^c$
• The fifth equality uses the rule $a^{-b} = 1/a^b$
• The final equality uses the special rule mentioned, $x = e^{\ln(x)}$