How do you find the integral of ${log}_{8} (2 x + 1) d x$ $log_8 (2x+1) dx$ ?

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Answer 1 · 2015-04-15T17:16:40

$\frac{1}{ln 8}$ $1/ln8$ [ x ln(2x+1) -x + $\frac{1}{2}$ $1/2$ ln(2x+1)] +C

To start with, change this to natural log. It would be $\frac{1}{ln 8}$ $1/ln8$ ln(2x+1)

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

The integral would be $\frac{1}{ln 8}$ $1/ln8$ [ln(2x+1) x - $\int$ $int$ $\frac{2}{2 x + 1}$ $2/(2x+1)$ xdx]

= $\frac{1}{ln 8}$ $1/ln8$ [x ln(2x+1) - $\int$ $int$ $\frac{2 x + 1 - 1}{2 x + 1}$ $(2x+1-1)/(2x+1)$ dx

= $\frac{1}{ln 8}$ $1/ln8$ [ x ln(2x+1) - $\int$ $int$ dx + $\int$ $int$ $\frac{1}{2 x + 1}$ $1/(2x+1)$ dx

= $\frac{1}{ln 8}$ $1/ln8$ [ x ln(2x+1) -x + $\frac{1}{2}$ $1/2$ ln(2x+1)] +C

How do you find the integral of log_8 (2x+1) dxlog8(2x+1)dxlog_8 (2x+1) dx?