1. ## Common Logarithmic Integral

Would anyone mind checking this to see if this looks correct? I'll explain more below:

$\displaystyle \int {\frac{{dx}} {{x\log _{10} x}}}$

Simple enough, I make a substitution:

$\displaystyle \begin{gathered} u = \log _{10} x \hfill \\ du = \frac{1} {{\ln 10}} \cdot \frac{1} {x}dx \hfill \\ dx = x\ln (10)du \hfill \\ \end{gathered}$

$\displaystyle \begin{gathered} \int {\frac{{x\ln (10)du}} {{x \cdot u}}} = \ln 10\int {\frac{1} {u}du = \ln 10\left[ {\ln |u| + c} \right]} = \hfill \\ \ln 10\left[ {\ln |\log _{10} x| + c} \right] = \ln 10 \cdot \ln |\log _{10} x| + C \hfill \\ \end{gathered}$

I'm not sure how you would simplify the final result any further.
This is what my book has for an answer. Am I still correct? Or did I go wrong somewhere?

$\displaystyle \ln 10 \cdot \ln |\ln x| + C$

Thank you.

2. Originally Posted by RedBarchetta
Would anyone mind checking this to see if this looks correct? I'll explain more below:

$\displaystyle \int {\frac{{dx}} {{x\log _{10} x}}}$

Simple enough, I make a substitution:

$\displaystyle \begin{gathered} u = \log _{10} x \hfill \\ du = \frac{1} {{\ln 10}} \cdot \frac{1} {x}dx \hfill \\ dx = x\ln (10)du \hfill \\ \end{gathered}$

$\displaystyle \begin{gathered} \int {\frac{{x\ln (10)du}} {{x \cdot u}}} = \ln 10\int {\frac{1} {u}du = \ln 10\left[ {\ln |u| + c} \right]} = \hfill \\ \ln 10\left[ {\ln |\log _{10} x| + c} \right] = \ln 10 \cdot \ln |\log _{10} x| + C \hfill \\ \end{gathered}$

I'm not sure how you would simplify the final result any further.
This is what my book has for an answer. Am I still correct? Or did I go wrong somewhere?

$\displaystyle \ln 10 \cdot \ln |\ln x| + C$

Thank you.
$\displaystyle \log_{10}x$ to $\displaystyle \frac {\ln x}{\ln 10}$ from the beginning. and then let $\displaystyle u = \ln x$. things work out slightly nicer and look more aesthetic, and you can get to your final answer without doing the algebraic manipulation on your second to last line