Question

How do you find the integral of `log_8 (2x+1) dx`?

Answer 1

`1/ln8` [ x ln(2x+1) -x + `1/2` ln(2x+1)] +C

To start with, change this to natural log. It would be `1/ln8` ln(2x+1)

Now integrate it by parts assuming 1 as the other function multiplied to ln(2x+1).

The integral would be `1/ln8`[ln(2x+1) x - `int` `2/(2x+1)`xdx]

= `1/ln8` [x ln(2x+1) - `int` `(2x+1-1)/(2x+1)` dx

= `1/ln8` [ x ln(2x+1) -`int` dx +`int` `1/(2x+1)` dx

=`1/ln8` [ x ln(2x+1) -x + `1/2` ln(2x+1)] +C