Evaluate: ∫(√cotx + √tanx)dx

Let I = ∫(√cotx + √tanx)dx ⇒ ∫(1+tanx)√tanxdx
Put tanx = z² ⇒ sec²xdx = 2zdz
dx = 2zdz/(1+z⁴)
I = ∫(1+z²)z.2zdz/(1+z⁴) = 2∫z²(1+z²)dz/(1+z⁴) = 2∫(1+1/z²)/(z² + 1/z²)dz = 2∫(1+1/z²)((z - 1/z)² + 2)dz
Put z - 1/z = u ⇒ (1+1/z²)dz = du
I = 2∫du/(u² + (√2)²)
I = 2.(1/√2)tan⁻¹(u/√2) + C = √2tan⁻¹((z - 1/z)/√2) + C = √2tan⁻¹((tanx - 1/tanx)/√2) + C = √2tan⁻¹((tanx - cotx)/√2) + C