see also the docstring of replace for instructions on how it might be used
expr=1/(1-x)+1/(1+x)
i=Integral(expr,x)
i.replace(lambda arg: arg.is_Add, lambda arg: arg.together().expand())
Integral(2/(-x**2 + 1), x)
On Monday, October 27, 2014 9:57:49 AM UTC-5, Francesco Bonazzi wrote:
Interesting, I didn't know of epath, it looks like it supports
type-matching, which current wildcards do not support.
On Friday, October 24, 2014 4:50:16 PM UTC+2, Mateusz Paprocki wrote:
Hi,
On 24 October 2014 15:16, Francesco Bonazzi franz@gmail.com
javascript: wrote:
Consider
Consider this use case
In [97]: expr = 1/(1-x) + 1/(1+x)
In [98]: e2 = Integral(expr, x)
In [99]: e2
Out[99]:
⌠
⎮ ⎛ 1 1 ⎞
⎮ ⎜─ + ──⎟ dx
⎮ ⎝x + 1 -x + 1⎠
⌡
Suppose now I want to act on the expression inside the integral by
Hi,
On 24 October 2014 15:16, Francesco Bonazzi franz.bona...@gmail.com wrote:
Consider this use case
In [97]: expr = 1/(1-x) + 1/(1+x)
In [98]: e2 = Integral(expr, x)
In [99]: e2
Out[99]:
⌠
⎮ ⎛ 1 1 ⎞
⎮ ⎜─ + ──⎟ dx
⎮ ⎝x + 1 -x + 1⎠
⌡
Suppose now I want to act on
