dy/dx =, then y =

Question

I dy/dx = (-1)/sqrt(1-x^2) + 1/x , then y =

A. sin^-1 x - 1/x^2 + C

B. cos^-1 x - 1/x^2 + C

C. sin^-1 x - ln |x| + C

D. cos^-1 x + ln |x| + C

Answer 1

Integral of 1/sqrt(1-x^2) = arcsin(x). The -1 goes outside, so it's -arcsin(x)

Integral of 1/x = lnx

Add a constant.

-arcsin(x) + ln|x| + C = y