Integrate 3x4 x243x please show steps I need to learn thisSo

Solution

integral (4+3 x)/((3-x) (4+x^2)) dx For the integrand (3 x+4)/((3-x) (x^2+4)), cancel common terms in the numerator and denominator: = integral (-3 x-4)/((x-3) (x^2+4)) dx For the integrand (-3 x-4)/((x-3) (x^2+4)), use partial fractions: = integral (x/(x^2+4)-1/(x-3)) dx Integrate the sum term by term and factor out constants: = integral x/(x^2+4) dx- integral 1/(x-3) dx For the integrand x/(x^2+4), substitute u = x^2+4 and du = 2 x dx: = 1/2 integral 1/u du- integral 1/(x-3) dx The integral of 1/u is log(u): = (log(u))/2- integral 1/(x-3) dx For the integrand 1/(x-3), substitute s = x-3 and ds = dx: = (log(u))/2- integral 1/s ds The integral of 1/s is log(s): = (log(u))/2-log(s)+constant Substitute back for s = x-3: = (log(u))/2-log(x-3)+constant Substitute back for u = x^2+4: = 1/2 log(x^2+4)-log(x-3)+constant