# Thread: Common Logarithmic Integral

1. ## Common Logarithmic Integral

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

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

Simple enough, I make a substitution:

$
\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}
$

$
\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?

$
\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:

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

Simple enough, I make a substitution:

$
\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}
$

$
\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?

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

Thank you.
you're good. your final answer is correct

your methodology is slightly weird to me though. i would have just changed
$\log_{10}x$ to $\frac {\ln x}{\ln 10}$ from the beginning. and then let $u = \ln x$. 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