how do you integrate ln(x^2-64)?

Answer 1

First, break the function apart using properties of logarithms.
int[ln(x-8)] + int[ln(x+8)]

Next, you have to get the x-8 in a different form. Do a u substitution on x-8 to get a clean change of variables. Similarly, do a different u sub on x+8

int[ln(u)*du] + int[ln(w)*dw]

Now, to integrate a logarithm function you have to use integration by parts. I will just do the first one and then the second one will be obvious.

Let y=ln(u) and dz=du
So dy=du/u and z=u
The formula is yz-int[z*dy]
So we get u*ln(u) - int[u*du/u] = u*ln(u) - int(du) = u*ln(u) - u.

Now resubstitute what u is.

(x-8)*ln(x-8) - (x-8)

Alternatively, for the second part, you should get something similar. The final answer, with the constant of integration, is:

(x-8)*ln(x-8) - (x-8) + (x+8)*ln(x+8) - (x+8) + C

And simplifiy to get:

(x-8)*ln(x-8) + (x+8)*ln(x+8) -2x + C


Viola!

Answer 2

You have ln(x^2)/ln 64 according to logarithmic properties.

The 1/(ln 64) comes out of the integral and one integrates ln(x^2) dx. Whatever the integral of ln(x^2) is - divide by ln 64.

Answer 3

By making them go to the same schools and ride the same buses.

Answer 4

Hahahahahaha! "Serious Mandy," that was funny.