∫01tan−1(1−x+x2)dx=

∫01tan−1(1−x+x2)dx=∫01π2−cot−1(1−x+x2)dx                                          =π2−∫01tan−111−x+x2dx=π2-∫01tan−1x+(1−x)1−x(1−x)dx                                         =π2−∫01(tan−1x+tan−1(1−x))dx                             =π2−∫01tan−1x−∫01tan−1(1−x)dx,x→1−x=π2−2∫01tan−1xdx                                         =π2−2[xtan−1x|01+2∫01x1+x2dx                                         =∫012x1+x2dx=ln(1+x2)|01=ln2