what is the integral of (x^3 - 2x^2 + x + 1) / ( x^4 + 5x^2 + 4)?

Answer 1

You can use partial fractions method for this.

x^4 + 5x^2 + 4 factors into (x^2 + 1) (x^2 + 4)

now you want to see how you can split the original, large fraction, into something more managable in the form:
something / (x^2+1) + something/(x^2+4)
because those two separate fractions you can do integrals for.

the "something" would come in some form Ax+B, where A and B can be any coefficient, including 0. So now you have to solve for the coefficients:

(Ax+B) / (x^2+1) + (Cx+D) / (x^2+4) = (x^3 - 2x^2 + x + 1) / ( x^4 + 5x^2 + 4)
so multiply out the denominators...

(Ax+B)(x^2+4) + (Cx+D)(x^2+1) = x^3 - 2x^2 + x + 1
Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Cx + Dx^2 + D = x^3 - 2x^2 + x + 1
x^3 (A+C) + x^2(B+D) + x(4A+C) + (4B+D) = x^3 - 2x^2 + x + 1

from these, we just do system of equations:
A+C = 1 -->C = 1-A
4A+C = 1 --->4A + 1-A = 1, A = 0, C = 1

B+D = -2 -->D = -2-B
4B+D = 1
4B - 2-B = 1
3B = 3, B = 1, D = -3

so the new integral can be written as
int( (1/(x^2+1) + (x-3)/(x^2+4) dx)
int((1/(x^2+1) + x/(x^2+4) - 3/(x^2+4) dx)

now you can split up each and do them individually,
int(1/(x^2+1) = arctanx

int(x/(x^2+4)) , u = x^2+4, du = 2xdx
1/2 int(1/u du)
= 1/2 ln(x^2 + 4)

-3 int(1/x^2+4) = -3/2 arctan(x/2)


so the final answer is:
arctanx + 1/2 ln(x^2+4) - 3/2 arctan(x/2)

Answer 2

The answer is actually arctan(x) + 1/2 ln(x^2+4) + 3/2 arctan(2/x)
Note that it's arctan(2/x), not arctan(x/2)