Integrate 1/x+ 2/x^3+2

Answer 1

To integrate 1/(x + 2/x^3 + 2), we need to first simplify the denominator.

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

Now, we can integrate:

∫1/(x + 2/x^3 + 2) dx

Let u = x + 2
du = dx

Replacing x + 2 with u, we get:

∫1/(u + 1/u^3) du

= ∫u^(-3)/(u^4 + 1) du

Now we have a rational function, which can be integrated by partial fractions.

(u^(-3))/(u^4 + 1) = A/(u - i) + B/(u + i) + Cu/(u^2 + 1) + Du/(u^2 + 1)^2

Multiplying through by the denominator and simplifying, we get:

1 = A(u^3 + u) + B(u^3 - u) + Cu^2 + Du

We can solve for A, B, C, and D by comparing coefficients.

After finding the values of A, B, C, and D, you can integrate each term separately to get the final result. Let me know if you need more help with this calculation.