rational integral

[AntonioLao]

The integral of the rational function given by x/(1+x) is x-log(1+x) and the integral of the rational function given by x/(1-x) is –log(x-1)-x. Consequently, the power series expansion of log(1+x) is expressed using continued fraction as x(1-x(1/2-x(1/3-x(1/4-x(1/5-…))))) for all |x|<1. In these cases, the base of the log function is given as Euler’s number. The graph of x-log(1+x) in the real domain is the open set (-1, infinity) since if x=-1 then x-log(1+x) is meaningless in the real domain. However, x=-1 is the one and only vertical asymptote of x-log(1+x) such that as x approaches -1, x-log(1+x) approaches positive infinity. Nonetheless, if x=0 then x-log(1+x)=0 or simply means that x=log(1+x) as the singular solution of the rational integral where and when both x and x-log(1+x) are equal.