Can This Integral be Integrated?

[c0der]

Homework Statement

∫( log(1 - x) / (x + 1), x )

I have tried integrating this and had no luck.

Homework Equations

As above

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

 

[Dick]

Homework Statement

∫( log(1 - x) / (x + 1), x )

I have tried integrating this and had no luck.

Homework Equations

As above

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

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

 

[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.

 

[ehild]

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

ehild

 

[LCKurtz]

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

 

[Giant]

Is it not ∫[log(1−x/x+1)]dx?

Isn't that just separation?

I interpret it as ∫[log(1−x)]/[1+x] dx

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

 

[c0der]

Thank you, it's as LCKurtz interpreted it. There is no hope solving this second order system of ODEs with varible coefficients via the Peano Baker method as the series is almost impossible to represent via elementary functions.
This is just one of the integrals I got in the series:

http://integrals.wolfram.com/index....+-+log((b+2*t*(L-x))/(b+2*t*L)))&random=false

 

[Saitama]

Thank you, it's as LCKurtz interpreted it. There is no hope solving this second order system of ODEs with varible coefficients via the Peano Baker method as the series is almost impossible to represent via elementary functions.
This is just one of the integrals I got in the series:

http://integrals.wolfram.com/index....+-+log((b+2*t*(L-x))/(b+2*t*L)))&random=false

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

 

[c0der]

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

 

[Saitama]

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

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?

 

[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