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

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Example 22 Find (i) β«1βπ^π₯ (tan^(β1)β‘π₯+ 1/(1 + π₯^2 )) ππ₯ β«1βγπ^π₯ [tan^(β1)β‘γπ₯+1/(1+π₯^2 )γ ]ππ₯γ Putting π(π₯)=tan^(β1)β‘π₯ π^β² (π₯)= 1/(1 + π₯^2 ) Replacing π(π₯) by (tan^(β1)β‘π₯ ) and π^β² (π₯) by 1/(1 +γ π₯γ^2 ), we get β«1βγπ^π₯ [tan^(β1)β‘γπ₯+1/(1+π₯^2 )γ ]ππ₯=π^π₯ [tan^(β1)β‘π₯ ]+πΆγ =π^π₯ tan^(β1)β‘γπ₯+πΆγ Example 22 Find (ii) β«1β((π₯^2 + 1) π^π₯)/(π₯ + 1)^2 ππ₯ Integrating β«1βγ(π₯^2 + 1)/(π₯+1)^2 .π^π₯ ππ₯γ =β«1βγ(π₯^2+ 1 + 1 β 1)/(π₯+1)^2 .π^π₯ .ππ₯γ =β«1βγ(π₯^2 β 1 + 1 + 1)/(π₯+1)^2 .π^π₯ .ππ₯γ =β«1βγ[(π₯^2β1)/(π₯+1)^2 +2/(π₯+1)^2 ] π^π₯ ππ₯γ =β«1βππ₯[(π₯^2β(1)^2)/(π₯+1)^2 +2/(π₯+1)^2 ]ππ₯ =β«1βππ₯[(π₯β1)(π₯+1)/(π₯+1)^2 +2/(π₯+1)^2 ]ππ₯ =β«1βππ₯[(π₯β1)(π₯+1)/(π₯+1)^2 +2/(π₯+1)^2 ]ππ₯ =β«1βππ₯[(π₯ β 1)/(π₯ + 1)+2/(π₯+1)^2 ]ππ₯ Putting π(π₯)=(π₯ β 1)/(π₯ + 1) β΄ π^β² (π₯)=π/ππ₯ [(π₯ β 1)/(π₯ + 1)] Here, β π’(π₯)=π₯β1 π^β² (π₯)=(1.(π₯ + 1) β1 (π₯ β 1))/(π₯ + 1)^2 π^β² (π₯)=(π₯ + 1 β π₯ + 1)/(π₯ + 1)^2 π^β² (π₯)=2/(π₯ + 1)^2 β΄ Replacing π(π₯) by (π₯ β 1)/(π₯ + 1) and π^β² (π₯) by 2/(π₯ + 1)^2 , we get β«1βγ(π₯^2 + 1)/(π₯ + 1)^2 .ππ₯=β«1βππ₯[(π₯ β 1)/(π₯ + 1)+2/(π₯ + 1)^2 ]ππ₯γ =π^π₯ [(π₯ β 1)/(π₯ + 1)]+πΆ =(π₯ β 1)/(π₯ + 1).π^π₯+πΆ

Class 12
Important Question for exams Class 12