# Homework Help: Simple Integral Question

1. Sep 1, 2010

### mohabitar

A little rusty on this stuff.

Integral of: 2x/2x+1 dx

What method would I use here?

2. Sep 1, 2010

### ehild

2x/2x+1 = 1+1 =2. Integral of 2 is 2x+c.

Or you meant 2x/(2x+1)? If so, you have to use parentheses.
Than use the substitution 2x+1=y.

ehild

3. Sep 1, 2010

### mohabitar

Sorry yes this is what I meant: 2x/(2x+1)

However, if I use u=2x+1, du is 2dx, and this is not present on the top. 2x is. If I solve for dx, I get du/2. This still doesnt help out much. Can you clarify some more please..

4. Sep 2, 2010

### ehild

u=2x+1, du =2dx, dx=du/2.

$$\int\frac{2x}{2x+1}dx =\frac{1}{2} \int\frac{u-1}{u}dx$$

Can you proceed from here?

ehild

5. Sep 2, 2010

### mohabitar

Sorry this still doesnt help narrow it down. There are still variables in the numerator and denom, so I cant just solve this right away. This looks familiar though. I think I have to break this up into two integrals, but I dont have the slightest memory of how to do that.

6. Sep 2, 2010

### ehild

Divide the numerator by the denominator.

ehild

7. Sep 2, 2010

### mohabitar

Ah ok looks like I'll have to go reteach myself long division :/

Is this the only way this can be solved? How about splitting it into 2 integrals, or was I wrong about that?

8. Sep 2, 2010

### ehild

No, you split it into two integrals after you did that division. It is simple, just think, how to simplify (2x+6)/2, for example?

ehild

9. Sep 2, 2010

### HallsofIvy

If you have difficulty with long division (and you really shouldn't by the time you are taking Calculus), then divide after the substitution:
$$\int\frac{2x}{2x+1}dx= \frac{1}{2}\int\frac{u- 1}{u}du=\frac{1}{2}\int \frac{u}{u}+ \frac{1}{u} du= \frac{1}{2}\int(1- \frac{1}{u})du= \frac{1}{2}\int(1- u^{-1})$$