Помогите пожалуйста решить, хоть что-то по математическому анализу, очень прошу кто разбирается напишите понятное решение.

Take the integral:
integral (x^3+5)/(x^2-2 x-3) dx
For the integrand (x^3+5)/(x^2-2 x-3), do long division:
= integral (x+8/(x-3)-1/(x+1)+2) dx
Integrate the sum term by term and factor out constants:
= 2 integral 1 dx+8 integral 1/(x-3) dx+ integral x dx- integral 1/(x+1) dx
For the integrand 1/(x-3), substitute u = x-3 and du = dx:
= 8 integral 1/u du+2 integral 1 dx+ integral x dx- integral 1/(x+1) dx
For the integrand 1/(x+1), substitute s = x+1 and ds = dx:
= - integral 1/s ds+8 integral 1/u du+2 integral 1 dx+ integral x dx
The integral of 1/s is log(s):
= -log(s)+8 integral 1/u du+2 integral 1 dx+ integral x dx
The integral of 1/u is log(u):
= -log(s)+8 log(u)+2 integral 1 dx+ integral x dx
The integral of 1 is x:
= -log(s)+8 log(u)+2 x+ integral x dx
The integral of x is x^2/2:
= -log(s)+8 log(u)+x^2/2+2 x+constant
Substitute back for s = x+1:
= 8 log(u)+x^2/2+2 x-log(x+1)+constant
Substitute back for u = x-3:
= x^2/2+2 x+8 log(x-3)-log(x+1)+constant
Factor the answer a different way:
= 8 log(x-3)+1/2 (x (x+4)-2 log(x+1))+constant
Which is equivalent for restricted x values to:
Answer: |
| = x^2/2+2 x+8 log(3-x)-log(x+1)+constant