∫01Tan−1(2x1–x2)dx=
π2−ln 2
π−ln 2
π+ln 2
π4
∫01Tan−1(2x1−x2)dx=∫012Tan−1xdx
=2[(Tan−1x.x)01]−2∫0111+x2.xdx
=2(π4−0)−∫012x1+x2dx
=π2−(log(1+x2))01
=π2−log2