Find the partial fraction of 2/((x^2+2)^2)

Question

Partial fraction

Answer 1

I'm guessing that you're trying to integrate and you're thinking you can use partial fractions to simplify the integration. But 2/(x²+2)² is already as reduced as it can be.

If you're trying to integrate, then you'll need to use substitution, starting with x=(tanθ)√2.

If integration is what you're trying to do, add a comment to your question, and I think I can answer it.

Comments

x=(tanθ)√2, dx=(sec²θ)√2dθ, x²+2=2sec²θ.

Integrand 2dx/(x²+2)²=

(2/(4sec⁴θ))(√2(sec²θ)dθ=(1/√2)cos²θdθ=(1/√2)(½(cos(2θ)+1))dθ.

Integrating: (1/(2√2))(sin(2θ)/2+θ)=(1/(2√2))(sinθcosθ+θ)+C.

If tanθ=x/√2, sinθ=x/√(x²+2), cosθ=√2/√(x²+2), sinθcosθ=x√2/(x²+2).

∫2dx/(x²+2)²=(1/(2√2))(x√2/(x²+2)+tan^-1(x/√2))+C=½x/(x²+2)+(1/(2√2))tan^-1(x/√2)+C.