1. ## Integration

Hello all~ I'm doing a problem where Integration by parts is used and i have reached a part where i'm stumped.

The problem is

Integrate ln(2x+1)dx

I've let u = ln(2x+1) dv = dx

du = 2 / (2x+1)dx v = x

xln(2x+1) - integrate (2x)/(2x + 1) dx

I'm having problems integrating (2x) / (2x + 1) dx. It seems something so simple but when I tried it out a couple times, it simply doesn't match up to my teachers answer. He was able to make the above into

Integrate 1 - (1/(2x + 1)) which I can do fairly easily by using substitution. Any help is appreciated.

2. Hello,

I'm having problems integrating (2x) / (2x + 1) dx. It seems something so simple but when I tried it out a couple times, it simply doesn't match up to my teachers answer. He was able to make the above into
$\displaystyle \frac{2x}{2x+1}=\frac{2x+1}{2x+1}-\frac{1}{2x+1}=1-\frac{1}{2x+1}$

Does it help ?

3. Originally Posted by JonathanEyoon
Hello all~ I'm doing a problem where Integration by parts is used and i have reached a part where i'm stumped.

The problem is

Integrate ln(2x+1)dx

I've let u = ln(2x+1) dv = dx

du = 2 / (2x+1)dx v = x

xln(2x+1) - integrate (2x)/(2x + 1) dx

I'm having problems integrating (2x) / (2x + 1) dx. It seems something so simple but when I tried it out a couple times, it simply doesn't match up to my teachers answer. He was able to make the above into

Integrate 1 - (1/(2x + 1)) which I can do fairly easily by using substitution. Any help is appreciated.

rewrite as follows (or use long division)

$\displaystyle \frac{2x}{2x+1}=\frac{2x+1-1}{2x+1}=\frac{2x+1}{2x+1}+\frac{-1}{2x+1}=1-\frac{1}{2x+1}$

I hope this helps.

Edit: I'm too slow Go $\displaystyle {\mu}$

4. I'm sorry this is kinda embarrassing since it's algebra but could you tell me step by step why it's reduced like that? I remember vaguely but it was so long ago