bigubau said:You're doing something wrong. Just write the whole polynomial in y and equal it to 1. You should get a system of 4 equations with 4 unknowns which is very easy to solve.
Out of curiosity, why did you switch from x that was used in your integral to y here?n4rush0 said:So I set
1/[y(y-1)(y^2+y+1)] = A/y + B/(y-1) + (Cy+D)/(y^2+y+1)
1 = (y-1)(y^2+y+1)A + y(y^2+y+1)B + y(y-1)(Cy+D)
y = 0, A = -1
y = 1, B = 1/3
Not sure what to do for C and D
The general formula for integrating 1/(x^4 - x) is ∫1/(x^4 - x)dx = 1/3ln|x^3 - 1| + C.
The substitution method for solving the integral of 1/(x^4 - x) involves substituting u = x^2 - 1, du = 2xdx, and rewriting the integral as ∫1/(u^2 + 1)du.
To solve the integral of 1/(x^4 - x) using partial fractions, we first use long division to rewrite the integrand as 1/(x^4 - x) = 1/(x(x^3 - 1)) = 1/(x(x-1)(x^2 + x + 1)). Then, we can use the method of partial fractions to decompose the rational function into simpler fractions, and integrate each part separately.
Yes, the integral of 1/(x^4 - x) can be solved using trigonometric substitution. We can use the substitution x = tanθ to rewrite the integral as ∫1/(tanθ(tan^3θ - tanθ))sec^2θdθ, and then use the trigonometric identity tan^2θ = sec^2θ - 1 to simplify the integrand.
Yes, there are other methods for solving the integral of 1/(x^4 - x), such as using the Weierstrass substitution or the Euler substitution. However, these methods may not always be straightforward and may require a deeper understanding of advanced calculus concepts.