In a partial fraction decomposition (p.f.d.), you first factor the denominator as far as possible, i.e., into linear factors, complete squares, and irreducible quadratic factors. In your case, the denominator has an obvious root at x=-1, so you can factor out the linear term (x+1) and are left with an irreducible quadratic:
x³-x²+x+3 = (x+1)(x²-2x+3)
According to the rules, the partial fractions must then be of the form
A/(x+1) + (Bx+C)/(x²-2x+3)
for constants A, B, and C, which we need to determine. This you do by adding the two partial fractions:
(A(x²-2x+3)+(Bx+C)(x+1))/(x+1)(x²-2x+3)
Now compare the numerator of this fraction with the numerator of your original expression. They must be equal:
x²+5 = A(x²-2x+3)+(Bx+C)(x+1)
Even though this is one polynomial equation, it can be turned into 3 linear equations for A, B, and C. There are two ways of getting these linear equations:
1) since the relation is true for any x, you can pick "nice" values for x, such as 0,-1, and 1.
2) since all terms with an x², with an x, and with no x are independent of each other, you can extract their respective coefficients.
I will use method 1):
For x=-1 we get
6= 6A, implying A=1
For x=0, we get
5 =3A+C=3+C, implying C=2
For x=1, we get
6= 2A+2(B+C)= 6+2B, implying B=0.
Therefore, the p.f.d. is
(x²+5)/(x³-x²+x+3) = 1/(x+1) + 2/(x²-2x+3)
Now you can integrate both terms easily:
The first integral is obviously ln(x+1)+c.
For the second one you need to complete the square in the denominator ((x-1)²+2) and get the integral, after a substitution (u=x-1), √2 arctan((x-1)/√2)+c.
