1. Class 12
2. Important Question for exams Class 12

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Ex 7.4, 21 (π₯ + 2)/β(π₯^2 + 2π₯ + 3) β«1β(π₯ + 2)/β(π₯^2 + 2π₯ + 3) ππ₯=1/2 β«1β(2π₯ + 4)/β(π₯^2 + 2π₯ + 3) ππ₯ =1/2 β«1β(2π₯ + 2 + 4 β 2)/β(π₯^2 + 2π₯ + 3) ππ₯ =1/2 β«1β(2π₯ + 2)/β(π₯^2 + 2π₯ + 3) ππ₯+2/2 β«1βππ₯/β(π₯^2 + 2π₯ + 3) ππ₯ =1/2 β«1β(2π₯ + 2)/β(π₯^2 + 2π₯ + 3) ππ₯+β«1βππ₯/β(π₯^2 + 2π₯ + 3) ππ₯ Solving π°π I1=1/2 β«1β(2π₯ + 2)/β(π₯^2 + 2π₯ + 3) . ππ₯ Let π₯^2 + 2π₯ + 3=π‘ Diff both sides w.r.t.x 2π₯+2+0=ππ‘/ππ₯ ππ₯=ππ‘/(2π₯ + 2) Now, our equation becomes I1=1/2 β«1β(2π₯ + 2)/β(π₯^2 + 2π₯ + 3) . ππ₯ Putting the value of (4π₯βπ₯^2 ) and ππ₯ I1=1/2 β«1β(2π₯ + 2)/βπ‘ . ππ₯ I1=1/2 β«1β(2π₯ + 2)/βπ‘ . ππ‘/(2π₯ + 2) I1=1/2 β«1β1/βπ‘ . ππ‘ I1=1/2 β«1β1/(π‘)^(1/2) . ππ‘ I1=1/2 β«1β(π‘)^((β 1)/2) . ππ‘ I1=1/2 γπ‘ γ^((β1)/2 + 1)/((β1)/2 + 1) +πΆ1 I1= π‘ ^(1/2 )+πΆ1 I1= βπ‘+πΆ1 I1=β(π₯^2+2π₯+3)+πΆ Solving π°π I2=β«1β1/β(π₯^2 + 2π₯ + 3) . ππ₯ I2=β«1β1/β(π₯^2 + 2(π₯)(1) + 3) . ππ₯ I2=β«1β1/β(π₯^2 + 2(π₯)(1) β (1)^2 + (1)^2 + 3) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 β (1)^2 + 3) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 β 1 + 3) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 + 2) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 +(β2 )^2 ) . ππ₯ I2=β«1β1/β(π₯^2 + 2(π₯)(1) β (1)^2 + (1)^2 + 3) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 β (1)^2 + 3) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 β 1 + 3) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 + 2) . ππ₯ I2=β«1β1/β((π₯ + 1)^2 +(β2 )^2 ) . ππ₯ I2=πππβ‘|π₯+1+β((π₯ + 1)^2+(β2 )^2 )|+πΆ2 I2=πππβ‘|π₯+1+β(π₯^2+2π₯+1+2)|+πΆ2 I2=πππβ‘|π₯+1+β(π₯^2+2π₯+3)|+πΆ2 Putting the values of I1 and I2 in (1) β«1βγ(π₯ + 2)/β(π₯^2 + 2π₯ + 3).γ . ππ₯ =1/2 β«1β(2π₯ + 2)/β(π₯^2 + 2π₯ + 3) . ππ₯+β«1β1/β(π₯^2 + 2π₯ + 3) . ππ₯ =β(π₯^2+2π₯+3)+πΆ1+πππβ‘|π₯+1+β(π₯^2+2π₯+3)|+πΆ2 =β(π^π+ππ+π)+πππβ‘|π+π+β(π^π+ππ+π)|+πͺ

Class 12
Important Question for exams Class 12