how do i take the integral of the following function?

Question

find the integral from negative 3 to positive 3 for the function [(20/(1+x^2)) - 2 dx

Answer 1

D=∫[(20/(1+x²))-2]dx for -3≤x≤3. This is the area between the curve and the x-axis because the expression is zero when x=-3 or 3, so the extremes of the integral are at (-3,0) and (3,0), and the whole area is above the x-axis.

D=[20tan^-1(x)-2x]_-3³=(20tan^-1(3)-6)-(20tan^-1(-3)+6)=

20tan^-1(3)-6-(-20tan^-1(3)+6)=20tan^-1(3)-6+20tan^-1(3)-6=

40tan^-1(3)-12=37.9618 approx.