∫01 tan−1⁡1x2−x+1 is equal to

∫01 tan−1⁡1x2−x+1dx=∫01 tan−1⁡x−(x−1)1+x(x−1)=∫01 tan−1⁡xdx−∫01 tan−1⁡(x−1)dx=∫01 tan−1⁡xdx+∫01 tan−1⁡(1−1+x)dx=∫01 tan−1⁡xdx+∫01 tan−1⁡xdx=2∫01 tan−1⁡xdx=2xtan−1⁡x−∫x1+x2dx01=2xtan−1⁡x−12log⁡1+x201=21tan−1⁡1−12log⁡(2)−0−12log⁡1=2π4−12log⁡2−0=π2−log⁡2