The integral ∫xcos⁻¹((1-x²)/(1+x²))dx is equal to :(Note :(x>0))

∫xcos⁻¹((1-x²)/(1+x²))dx (x>0)
x=tanθ
dx=sec²θdθ
I=∫tanθcos⁻¹(cos2θ)sec²θdθ
I=∫2θtanθsec²θdθ
Using by parts ∫u.v dx=u∫v dx -∫(du/dx∫v dx)dx
I=2θ∫tanθsec²θdθ -∫2∫tanθsec²θdθ
=2θtan²θ/2 -[∫(tan²θ+sec²θ)dθ].
sec²x=1+tan²x
=θtan²θ +θ-tanθ+c
=θ(1+tan²θ)-tanθ+c
Replacing θ by x, I=(tan⁻¹x)(1+x²)-x+c
=-x+(1+x²)tan⁻¹x+c