Find : ∫(x2+1)(x+2)x2+x+1dxAnswer: 53log∣x+2∣+51log(x2+1)+51tan−1x+CSolutionResolve into partial fractions:(x2+1)(x+2)x2+x+1=x+2A+x2+1Bx+C.Comparing coefficients gives A=53, B=52, C=51. Hence∫x+23/5dx+51∫x2+12x+1dx=53log∣x+2∣+51log(x2+1)+51tan−1x+C.53log∣x+2∣+51log(x2+1)+51tan−1x+CCBSE Class 12 Board · 2016 · Subjective