Evaluate the Integral: $\int \frac{\log_{10}\ x}{x}\ dx$ $\int \frac{\log_{10}\ x}{x}\ dx$
$du=x\ln10\ dx$
$\log_{10}x\  \ln10+ C$
Is this answer correct? If not what step should I take to convert the log into a term I can manipulate?  
 A: I encourage you to read this then try doing the rest yourself.
$$\int \frac{\log_{10}\ x}{x}\ dx$$
Recall that,
$$\log_a(b)={{\ln(b)} \over {\ln(a)}}$$
So,
$$\int \frac{\log_{10}\ x}{x}\ dx=\int {{\ln(x)} \over {x \cdot \ln(10)}} dx$$
Note that the derivative of $\ln(x)$ is ${1 \over x}$. Can you continue from here?
Hint: use u-substitution
Extra problem:
Try integrating ${{\log_4(x^{1/3})} \over x}$
A: $$\int { \frac { \log _{ 10 }{ x }  }{ x } dx } =\int { \log _{ 10 }{ x } d\ln { x } =\log _{ 10 }{ x } \ln { x } -\int { \frac { \ln { x }  }{ x\ln { 10 }  } +C= }  } \\ =\log _{ 10 }{ x } \ln { x } -\frac { 1 }{ \ln { 10 }  } \int { \ln { x } d\ln { x } +C= } \log _{ 10 }{ x } \ln { x } -\frac { 1 }{ \ln { 10 }  } \ln { x } +C=\\ =\ln { x\left( \log _{ 10 }{ x } -\frac { 1 }{ \ln { 10 }  }  \right) +C } $$
A: Notice, $$\int \frac{\log_{10}\ x}{x}\ dx$$  $$\int\frac{\log x dx}{x\log 10}$$ Now, let $\log x=t\implies \frac{dx}{x}=dt$, hence we have $$\frac{1}{log 10} \int t dt$$  $$=\frac{1}{\log 10}\left(\frac{t^2}{2}\right)+C$$ Substituting the value of $t=\log x$
  $$=\frac{(\log_{e}x)^2}{2\log 10}+C$$
A: $\int \frac{\log_{10}\ x}{x}\ dx$ 
but $\log_{10}x=\frac{lnx}{ln10}$
So we have $\int\frac{lnx}{xln10}dx$
$$\frac{1}{ln10}\int\frac{lnx}{x}dx$$
Let $u=lnx$, therefore $\frac{du}{dx}=\frac{1}{x}$
$$\frac{1}{ln10}\int\frac{u}{x}xdu$$
$$\frac{1}{ln10}\int udu$$
$$\frac{1}{ln10}\frac{u^2}{2}+C$$
$$\frac{1}{ln10}\frac{(lnx)^2}{2}+C$$
