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

Question

Cameron Alexander · Accepted Answer

#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

Ezekiel Alexander · Answer

undefined To find the integral of log base 8 of (2x + 1) with respect to x, you can use substitution. Let u = 2x + 1, then du/dx = 2. Rearrange this equation to solve for dx, giving you dx = du/2. Now substitute u and dx in terms of x into the integral. The integral becomes (1/2) ∫(log base 8 of u) du.

Next, recall that log base 8 of u can be expressed as ln(u)/ln(8). So, the integral becomes (1/2) ∫(ln(u)/ln(8)) du. Using properties of logarithms, ln(8) can be factored out of the integral. This leaves you with (1/2)(1/ln(8)) ∫ln(u) du.

Integrating ln(u) with respect to u gives u ln(u) - u + C, where C is the constant of integration.

Substitute back u = 2x + 1 to get the final result: (1/2)(1/ln(8))(2x + 1)(ln(2x + 1) - 1) + C.