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Math Help - Common Logarithmic Integral

  1. #1
    Member RedBarchetta's Avatar
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    Common Logarithmic Integral

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

    <br />
\int {\frac{{dx}}<br />
{{x\log _{10} x}}} <br />

    Simple enough, I make a substitution:

    <br />
\begin{gathered}<br />
  u = \log _{10} x \hfill \\<br />
  du = \frac{1}<br />
{{\ln 10}} \cdot \frac{1}<br />
{x}dx \hfill \\<br />
  dx = x\ln (10)du \hfill \\ <br />
\end{gathered} <br />

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


    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?


    <br />
\ln 10 \cdot \ln |\ln x| + C<br />

    Thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Would anyone mind checking this to see if this looks correct? I'll explain more below:

    <br />
\int {\frac{{dx}}<br />
{{x\log _{10} x}}} <br />

    Simple enough, I make a substitution:

    <br />
\begin{gathered}<br />
  u = \log _{10} x \hfill \\<br />
  du = \frac{1}<br />
{{\ln 10}} \cdot \frac{1}<br />
{x}dx \hfill \\<br />
  dx = x\ln (10)du \hfill \\ <br />
\end{gathered} <br />

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


    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?


    <br />
\ln 10 \cdot \ln |\ln x| + C<br />

    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
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