Calculus
Name:
Anonymous
2009-08-15 16:54
I've been trying to solve this integral, but I'm absolutely stuck on how to solve it.
\int_{}^{}dx/(2x^2-1)
is /sci/ able to help?
Name:
Anonymous
2009-08-15 18:23
Integrate 1/(2x^-1) you mean? Just use difference of two squares and partial fractions.
Name:
Anonymous
2009-08-15 18:33
Get your shithematics outta here, this is a board for popular science.
Name:
Anonymous
2009-08-17 17:46
Name:
Anonymous
2009-08-18 2:53
It's a logarithmic variant, use partial fractions. Indefinite is (Log[Sqrt[2]-2x]-Log[Sqrt[2]+2x])/(2Sqrt[2]).
Name:
Anonymous
2009-08-24 0:14
>>1
\int_{}^{}dx/(2x^2-1)