Find ∫(x2+1)(x2+2)2xdxAnswer: log(x2+2x2+1)+CSolutionPut t=x2, so dt=2xdx:∫(t+1)(t+2)dt.Partial fractions: (t+1)(t+2)1=t+11−t+21, solog(t+1)−log(t+2)+C=logt+2t+1+C.Restoring t=x2,log(x2+2x2+1)+CCBSE Class 12 Board · 2022 · Subjective