# Math Help - Integral question

1. ## Integral question

How do you do this?

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

Just substitute $t=\log_8(x)$

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

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

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

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

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

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

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

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

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

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

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