# Integral of ln(x)/(x+1) from 1 ot 0

1. May 17, 2010

### coki2000

Hello,
what is it's solvation.I tried to solve it by parts but i found undefinable term(from ln(0)).Please give me an advice.Thanks.

2. May 17, 2010

### Cyosis

This integral doesn't have a primitive in terms of elementary functions.

Edit: didn't notice it was a definite integral.

3. May 17, 2010

### Hurkyl

Staff Emeritus
That should be a clue that you've forgotten about improper integrals....

4. May 20, 2010

### coki2000

If i integrete this integral by parts

$$\int_{0}^{1}\frac{ln(x+1)}{x}dx$$

ln(x+1)=u ,1/(x+1)dx=du and 1/xdx=dv, lnx=v

$$\int_{0}^{1}\frac{ln(x+1)}{x}dx=ln(x+1)lnx\mid_{0}^{1}-\int_{0}^{1} \frac{lnx}{x+1}dx$$

This time integral of ln(x+1)/x from 0 to 1 is not integratable but answer of this is (pi^2)/6.

5. May 20, 2010

### Cyosis

You can integrate it just fine, however there does not exist a primitive in terms of elementary functions. You can evaluate this integral by using the power series of ln(x+1).

6. May 20, 2010

### guerom00

The message #4 is in contradiction with the title of the thread…
Which integral is it ?

7. May 20, 2010

### Gib Z

No it isn't. The post showed that each integral yields the other. And as Cyosis mentioned, using the Taylor Series for ln(1+x) centered at x=0, we can get a series representation of the integral, and it so happens in this case it simplifies to a simple combination of well known constants.