If ∫xx2−4x+8dx=Klog⁡x2−4x+8+tan−1⁡x−22+C then the value of K is

Completing the square, we getx2−4x+8=(x−2)2+4 so put x−2=u∫xx2−4x+8dx=∫u+2u2+4du=12∫2uu2+4du+2∫duu2+4=12log⁡u2+4+tan−1⁡u2+C=12log⁡x2−4x+8+tan−1⁡x−22+C.