What Method Should Be Used to Integrate 2x/(2x+1) dx?

[mohabitar]

A little rusty on this stuff.

Integral of: 2x/2x+1 dx

What method would I use here?

 

[ehild]

A little rusty on this stuff.

Integral of: 2x/2x+1 dx

What method would I use here?

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

 

[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 doesn't help out much. Can you clarify some more please..

 

[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

 

[mohabitar]

Sorry this still doesn't help narrow it down. There are still variables in the numerator and denom, so I can't just solve this right away. This looks familiar though. I think I have to break this up into two integrals, but I don't have the slightest memory of how to do that.

 

[ehild]

Divide the numerator by the denominator.ehild

 

[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?

 

[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

 

[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:
$$\int\frac{2x}{2x+1}dx= \frac{1}{2}\int\frac{u- 1}{u}du=\frac{1}{2}\int \frac{u}{u}+ \frac{1}{u} du= \frac{1}{2}\int(1- \frac{1}{u})du= \frac{1}{2}\int(1- u^{-1})$$