1. Integral question

How do you do this?

2. Hello,
Originally Posted by VkL
How do you do this?

Just substitute $\displaystyle t=\log_8(x)$

note that $\displaystyle \log_8(x)=\frac{\ln(x)}{\ln(8)}$ (may be useful for the dt)

3. I'm pretty sure that $\displaystyle \log$ stands for $\displaystyle \ln,$ most of books use this.

4. $\displaystyle \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx$

If you let $\displaystyle t=log_8x$, then $\displaystyle dt=\frac{1}{xln(8)}$.

Thus you have $\displaystyle \int_{2}^{8}\frac{ln(8)\,dt}{t^2}$.

Pull out the $\displaystyle ln(8)$ to get $\displaystyle ln(8)\int_{2}^{8}\frac{\,dt}{t^2} = ln(8)*-\frac{1}{t} = ln(8)*-\frac{1}{log_8x} = -\frac{ln(8)}{log_8x}$.

Plugging in $\displaystyle 8$ gives you $\displaystyle -\frac{ln(8)}{log_88} = -\frac{ln(8)}{1}=-ln(8)$ and plugging in $\displaystyle 2$ gives you $\displaystyle -\frac{ln(8)}{log_82} = -\frac{ln(8)}{\frac{1}{3}} = ln(8)*-3 = -3ln(8)$.

Subtracting gives: $\displaystyle -ln(8)+3ln(8) = 2ln(8) = 2ln(2^3) = 6ln(2)$.

Thus, $\displaystyle \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx = 6ln(2)$.

5. Originally Posted by Krizalid
I'm pretty sure that $\displaystyle \log$ stands for $\displaystyle \ln,$ most of books use this.
With a base ?
I don't think so ^^

Also, it's not most of books which use it ><