Evaluate : ∫(x2+1)(x2+3)2xdxAnswer: 21logx2+3x2+1+CSolutionPut z=x2, dz=2xdx:∫(z+1)(z+3)dz.Partial fractions: (z+1)(z+3)1=2(z+1)1−2(z+3)1, so21log∣z+1∣−21log∣z+3∣+C=21logz+3z+1+C.Restoring z=x2,21logx2+3x2+1+CCBSE Class 12 Board · 2011 · Subjective