, then Raymond B. answered • 03/31/22
Math, microeconomics or criminal justice
y = the integral of -1/(1-x^2)^(1/2) + 1/x
= -arcsinx + lnx + c where c = an unknown constant
use integral tables
#86 formula in Salas, Hille, Etgen's Calculus, one and several variables, 10th Edition
integral of du/sqr(a-u^2) let u=x and a=1
then integral = arcsinx, add a negative sign
without the integral tables, you have a nearly insoluble problem.
integral of 1/x or integral of x^-1 = lnx
derivative of arcsinx = 1/sqr(1-x^2)
derivative of lnx = 1/x except when x=0
usual rule is increase the exponent by 1 and divide by that new number which would mean x^0/0. but that's undefined. that exception leads to a new rule: integral of 1/x = lnx
