- Thread starter
- #1
DreamWeaver
Well-known member
- Sep 16, 2013
- 337
Here's an alternative solution to solving the integral:
\(\displaystyle \int_0^1\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx\)
[Original post: calculus - A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ - Mathematics Stack Exchange]
I'll post a link on there so the OP can see my work...
\(\displaystyle \int_0^1\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx\)
[Original post: calculus - A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ - Mathematics Stack Exchange]
I'll post a link on there so the OP can see my work...