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

Apr 15, 2015

$\frac{1}{\ln} 8$ [ x ln(2x+1) -x + $\frac{1}{2}$ ln(2x+1)] +C

To start with, change this to natural log. It would be $\frac{1}{\ln} 8$ ln(2x+1)

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

The integral would be $\frac{1}{\ln} 8$[ln(2x+1) x - $\int$ $\frac{2}{2 x + 1}$xdx]

= $\frac{1}{\ln} 8$ [x ln(2x+1) - $\int$$\frac{2 x + 1 - 1}{2 x + 1}$ dx

= $\frac{1}{\ln} 8$ [ x ln(2x+1) -$\int$ dx +$\int$ $\frac{1}{2 x + 1}$ dx

=$\frac{1}{\ln} 8$ [ x ln(2x+1) -x + $\frac{1}{2}$ ln(2x+1)] +C