# Can this be integrated?

1. May 3, 2014

### c0der

1. The problem statement, all variables and given/known data

∫( log(1 - x) / (x + 1), x )

I have tried integrating this and had no luck.

2. Relevant equations

As above

3. The attempt at a solution

By parts

u = 1/(x+1) , du = ln(x+1)dx
dv = log(1-x) dx , v = (x-1) * ( log(1-x) - 1 )

uv - int(v du) = 1/(x+1) * (x-1) * ( log(1-x) - 1 ) - ∫( (x-1)*(log(1-x) - 1) * log(x+1) dx )

Gets worse as I continue

2. May 3, 2014

### Dick

It's not an elementary integral you can compute that way. I think you need nonelementary functions like polylogarithms.

3. May 4, 2014

### abitslow

I don't understand what this means: "∫( log(1 - x) / (x + 1), x )" .... why is there a comma? where is dx ?
d(ln(y))/dy = 1/y is true.
d(1/y) = ln(y) is NOT.
if x >= 1, then log(1-x) is undefined. so why do you have an indefinite integral? not to mention 1/(x+1) when x= -1 ?
But I don't see how to solve it...I'd try trig functions...but its been way too long...Dick may be (and probably is, for all I know) exactly correct - its not elementary, IDK.

4. May 4, 2014

### ehild

Is it not $\int{\log\left(\frac{1-x}{x+1}\right)dx}$?

ehild

5. May 4, 2014

### LCKurtz

I interpret it as $\int \frac{\log(1-x)}{1+x}~dx$

6. May 4, 2014

### Giant

Isn't that just separation?

Looking how OP solved, this seems more correct

Even I tried it I couldn't do it.
Polylogarithms was correct when I tried to do it with Wolfram Integrator.
http://mathworld.wolfram.com/Polylogarithm.html

7. May 4, 2014

8. May 4, 2014

### Saitama

Are you sure that you were asked to calculate the anti derivative instead of some definite integral?

9. May 4, 2014

### c0der

It's a definite integral, so I've always calculated this by obtaining the anti-derivative and subbing in the limits of integration.

10. May 4, 2014

### Saitama

Then you should have posted the definite integral. You can not evaluate every definite integral by obtaining the anti-derivative, there is a different way to handle them. In any case, can you post the definite integral you are assigned to solve?

11. May 4, 2014

### c0der

$$\int_0^x \frac{1}{b + 2ts}\textrm{ln}\left(\frac{b+2t(L-s)}{b}\right)ds$$

That's one of the integrals I extracted from the Peano Baker series when attempting to solve a system of second order ODE with variable coefficients

Last edited by a moderator: May 4, 2014