Find $p>1$ that ${\int\limits^p_1}\frac{1}{x}\,\mathrm{d}x={\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x$ Find $p>1$ that $${\displaystyle\int\limits^p_1}\dfrac{1}{x}\,\mathrm{d}x={\displaystyle\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x$$

\begin{align*}
    &{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x=\ln\left(\mid x \mid \right) && \vert \ \text{general integral}
\end{align*}
$F_1(x)=\ln\left({\mid x \mid} \right)+C$
\begin{align*}
    &{\displaystyle\int}1\cdot\ln\left(x\right )\,\mathrm{d}x && \vert \ 2. \text{ with } f'=1, g=\ln(x)\\
    &=x\ln\left(x\right)-{\displaystyle\int}1\,\mathrm{d}x\\
    &=x\ln\left(x\right)-x
\end{align*}
$F_2(x)=x\ln\left(x\right)-x+C$

\begin{align*}
    \left[\ln\left(\mid{x}\mid\right)+C\right]^p_1&=\left[\ln({\mid p \mid})+C\right]-\left[\ln({\mid 1 \mid})+C\right]\\
    &=\ln\left(p\right)
\end{align*}
\begin{align*}
    &\left[x\ln\left(x \right)-x+C\right]^p_1=\left[p\ln\left(p \right)-p+C\right]-\left[1\ln\left(1\right)-1+C\right]
\end{align*}
It was already mentioned, that I had a typo. I corrected everything and contributed my own solution.
 A: $ \int_1^p \frac{1}{x} dx=\int_1^p \ln x dx \iff \ln p = p \ln p-p+1 \iff (1-p) \ln p=1-p \iff \ln p=1 \iff p=e.$
A: Find $p>1$ that
$$
    {\displaystyle\int\limits^p_1}\dfrac{1}{x}\,\mathrm{d}x={\displaystyle\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x
$$

\begin{align*}
    &{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x=\ln\left(\mid x \mid \right) && \vert \ \text{general integr}
\end{align*}
$F_1(x)=\ln\left({\mid x \mid} \right)+C$
\begin{align*}
    &{\displaystyle\int}1\cdot\ln\left(x\right )\,\mathrm{d}x && \vert \ 2. \text{ with } f'=1, g=\ln(x)\\
    &=x\ln\left(x\right)-{\displaystyle\int}1\,\mathrm{d}x\\
    &=x\ln\left(x\right)-x
\end{align*}
$F_2(x)=x\ln\left(x\right)-x+C$

\begin{align*}
    \left[\ln\left(\mid{x}\mid\right)+C\right]^p_1&=\left[\ln({\mid p \mid})+C\right]-\left[\ln({\mid 0 \mid})+C\right]\\
    &=\ln\left(p\right)
\end{align*}
\begin{align*}
    \left[x\ln\left(x \right)-x+C\right]^p_1&=\left[p\ln\left(p \right)-p+C\right]-\left[1\ln\left(1 \right)-1+C\right]\\
    &=\left[p\ln\left(p \right)-p+C\right]-[-1+C]\\
    &=p\ln\left(p \right)-p+1
\end{align*}
Solve  $\ln\left(p\right)=p\ln\left(p \right)-p+1$ for $p$:
\begin{align*}
    &\ln\left(p\right)=p\ln\left(p \right)-p+1&&\vert \ -(1-p+p\ln(p)\\
    &\iff-1+p+\ln(p)-p\ln(p)=0\\
    &\iff-(p-1)\cdot (\ln(p)-1)=0&& \vert \ \cdot (-1)\\
    &\iff (p-1)\cdot (\ln(p)-1)=0 && \vert \ p>1\\
    &\iff  (\ln(p)-1)=0\\
    &\iff (\ln(p))=1&& \vert \ \exp{()}\\
    &\iff \underline{\underline{p=e}}
\end{align*}
