The value of the integral ∫₁₀ x cot⁻¹(1-x²+x⁴)dx

I=∫₁₀xtan⁻¹(1/(x²+x²-1))dx
x²=t ⇒2xdx=dt
I=1/2∫₁₀(tan⁻¹t-tan⁻¹(t-1))dx=1/2∫₁₀(tan⁻¹t)dt -1/2∫₁₀(tan⁻¹(t-1))dt=∫₁₀(tan⁻¹t)dt
tan⁻¹t=θ ⇒t=tanθ
dt=sec²θdθ
∫π/4₀θ.sec²θdθ
I=|(θtanθ)|π/4₀ -∫π/4₀(tanθ)dθ=(π/4 -0)-ln|secθ|π/4₀=π/4 -(ln√2 -0)=π/4 -1/2ln2