Evaluate int (log10x)/(2x) dx

[confused_07]

Evaluate int (log10x)/(2x) dx

I m using the integral fomula:

int a^u du = (a^u)/(ln a) + C

So, I have so far (and I am stuck):

int (log10x)/(2x) dx
= int [(ln x)/(ln 3)] / 2x dx
= int (ln x - ln 3) / 2x

I can't figure out how to get it into the integral formula from here

 

[pka]

What does log10x mean?
Which of these is it? \(\displaystyle \log _{10} \left( {10x} \right),\quad \log _{10} \left( x \right),\quad \ln (10x)\)

In what you wrote where does ln(3) come from?

 

[confused_07]

Sorry, it's the middle one with 10 being the subscript (I didn't know how to type it).

ln3 came from the formula out of my text that says:

log_(a) * x
= [log_(e) * x] / [log_(e) * a]
= ln x / ln a

Sorry again, that was a typo... should be ln 10

 

[soroban]

Hello, confused_07!

Integrate: \(\displaystyle \L\: \int\frac{\log_{_{10}}(x)}{2x}\,dx\)

We have: \(\displaystyle \L\:\frac{1}{2}\int\frac{\log_{_{10}}(x)}{x}\,dx\)

Let: \(\displaystyle \,u\,=\,\log_{10}(x)\;\;\Rightarrow\;\;du\,=\,\frac{dx}{x\cdot\ln(10)}\;\;\Rightarrow\;\;dx\,=\,x\cdot\ln(10)\cdot du\)

Substitute: \(\displaystyle \;\frac{1}{2}\L\int\frac{u}{x}\)\(\displaystyle \cdot[x\cdot\ln(10)\cdot du] \;=\;\frac{1}{2}\cdot\ln(10)\L\int\)\(\displaystyle u\,du\;=\;\frac{1}{4}\cdot\ln(10)\cdot u^2\,+\,C\)

Back-substitute: \(\displaystyle \:\frac{1}{4}\cdot\ln(10)\cdot\left[\log_{_{10}}(x)\right]^2\,+\,C\)

 

[confused_07]

Thanks Soroban. Real quick, how does the 1/2 become 1/4? I understand everything else but that. (I am sure I'll feel stupid when you answer this... )

 

[skeeter]

Chief ...

\(\displaystyle \L \frac{1}{2} \ln{(10)} \int u du = \frac{1}{2} \ln{(10)} \cdot \frac{u^2}{2} + C = \frac{1}{4} \ln{(10)} \cdot u^2 + C\)

 

[confused_07]

Yep..... stupid!!!!! Where's that rock I crawl under.....


Dang, you said Chief? Most people don't know what a CPO is.....

 

[skeeter]

I'm an LCDR (Ret.)