int (x^3 - 6x - 20)/(x + 5) dx Find the indefinite integral.

The answer given by gsarora17 will simplify to x^3/3-(5x^2)/2+19x-115ln|x+5|+C
A direct method to this result is to simplify the integrand by long division or synthetic division and then integrating term by term:
int (x^3-6x-20)/(x+5)dx=int [ x^2-5x+19 - 115/(x+5)]dx
=x^3/3-(5x^2)/2+19x-115ln|x+5|+C as above.

int(x^3-6x-20)/(x+5)dx
Let's evaluate the integral by applying integral substitution,
Let u=x+5, =>x=u-5
du=dx
int(x^3-6x-20)/(x+5)dx=int((u-5)^3-6(u-5)-20)/udu 
=int((u^3-5^3-3u^2*5+3u*5^2)-6u+30-20)/udu
=int(u^3-125-15u^2+75u-6u+10)/udu
=int(u^3-15u^2+69u-115)/udu
=int(u^2-15u+69-115/u)du
Now apply the sum rule,
=intu^2du-int15udu-int115/udu+int69du
=intu^2du-int15udu-115int(du)/u+69intdu
Use the following common integrals,
intx^ndx=x^(n+1)/(n+1)
and int1/xdx=ln(|x|)
=u^3/3-15u^2/2-115ln|u|+69u
Substitute back u=x+5,
=(x+5)^3/3-15/2(x+5)^2-115ln|x+5|+69(x+5)
Add a constant C to the solution,
=(x+5)^3/3-15/2(x+5)^2+69(x+5)-115ln|x+5|+C