Integration- partial fraction form question.?

Question

I need help with a strategy to find the integral of:
1 / (x(x^4+1)) dx.

I was thinking maybe partial fractions.. but since x^4+1 can't be factored any farther, I'm not quite sure what form to put in the numerator (a, ax+b, ax^2+bx+c, ect). Any help would be much appreciated. Thanks!

Answer 1

Use partial fractions:

1/ (x(x^4+1))
= A/x + (Bx^3 + Cx^2 + Dx+E)/(x^4 + 1)

A(x^4+1) + Bx^4 + Cx^3 + Dx^2 + Ex = 1

Equate the powers of x:
x^0 : A = 1
x^1: E = 0
x^2: D = 0
x^3: C = 0
x^4: A+B = 0 ---> B = -1

1/(x(x^4+1))
= 1/x - x^3/(x^4+1)

∫1/(x(x^4+1)) dx
= ∫1/x - x^3/(x^4+1)
= ∫1/x - (1/4)*(4x^3)/(x^4+1)dx
= ln(x) - (1/4)ln(x^4+1) + c

Answer 2

you'll want to decompose it as
A/x + (Bx^3 + Cx^2 + Dx + E)/(x^4+1).

that's equal to
(Ax^4 + A + Bx^4 + Cx^3 + Dx^2 + Ex)/(x(x^4+1)),
so you'll get the equations
A+B = 0
C = 0
D = 0
E = 0
A = 1,
and the partial fraction decomposition is

1/x - x^3/(x^4+1).

in general, each term of your partial fraction form should be such that the polynomial in the numerator is one degree lower than the polynomial in the denominator.

Answer 3

The answers given so far are excellent. So let me
just add a bit:
x^4+1 can be factored as
(x²+√2x+1)(x²-√2x+1).
Now use this result to do your integral.

Answer 4

GODBLESS!!!