- #1

laker88116

- 57

- 0

So far I know that I need to use integration by parts, I let [itex]u= ln(2x+1) [/itex] and so [itex] du= \frac {dx}{2x+1} [/itex]. Also, I said [itex] dv= dx [/itex] and [itex] v=x [/itex].

So then plugging this into the equation for integration I get:

[tex] xln(2x+1) - \int \frac {2x}{2x+1}dx [/tex]

Then I determine that I need to do integration by parts again on the latter half of the function. So, for [itex]\int \frac {2x}{2x+1}dx [/itex], I let [itex]u= 2x [/itex] and so [itex] du= xdx [/itex]. Also, I said [itex] dv= \frac {dx}{2x+1} [/itex] and [itex] v= \frac {ln(2x+1)}{2} [/itex].

So then plugging this into the equation for integration I get:

[tex]\int \frac {2x}{2x+1}dx = xln(2x+1) - \int ln(2x+1)dx [/tex]

Now, I have like terms so I say that:

[tex]\int ln(2x+1)dx = xln(2x+1) - [xln(2x+1) - \int ln(2x+1)dx]. [/tex]

I am not sure where I made an error here. Any help is appreciated.