Hello, jeph!
How do you integrate:
. . . . . . . . 4u
. . ∫ -------------------- du
. . . (1 + u)²(1 + u²)
You need Partial Fractions . . .
. . . . . 4u . . . . . . . . . . A . . . . . .B . . . . .Cu + D
-------------------- . = . ------- + ---------- + ---------
(1 + u)²(1 + u²) . . . . 1 + u . . (1 + u)² . . 1 + u²
Once you try to write the partial fractions, you have to respect this rule:
Denominator Containing Repeated Linear Factors
If a linear factor is repeated n times in the denominator, there will be n corresponding partial fractions with degree 1 to n.
You have to have that because you have to have the same polynomial as a denominator. Your expresion, does not have the same degree as the initial denominator.