Question

Решите интеграл 2/(1-4x^2) пожалуйста!

Answer 1

$\displaystyle \int \dfrac2{1-4x^2}~dx=2\int\dfrac1{1-4x^2}~dx=\int\dfrac1{1-u^2}~du, ((u=2x)) = \dfrac12\cdot\ln\dfrac{u+1}{-u+1}+const=\dfrac12\ln\dfrac{2x+1}{1-2x}+const$