CBSE Class 12 Sample Paper for 2021 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Find β« x 2 + 1 / (x 2 + 2) (x 2 + 3) dx

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Question 33 Find β«1βγ(π₯^2 + 1)/((π₯^2 + 2) (π₯^2 + 3)) ππ₯γ Putting π^π=π (π₯^2 + 1 )/((π₯^2 + 2) (π₯^2 + 3) )=(π¦ + 1)/((π¦ + 2) (π¦ + 3) ) We can write this in form (π¦ + 1)/((π¦ + 2) (π¦ + 3) )=π΄/((π¦ + 2) ) + π΅/((π¦ + 3) ) (π¦ + 1)/((π¦ + 2) (π¦ + 3) )=(π΄(π¦ +3) + π΅ (π¦ + 2))/((π¦ + 2) (π¦ + 3) ) By cancelling denominator π¦+1=π΄(π¦ +3) + π΅ (π¦ + 2) Putting y = β3 β3+1=π΄(β3+3)+π΅(β3+2) β2=π΄ Γ 0+π΅ Γ β1 β2=βπ΅ π©=π Putting y = β2 β2+1=π΄(β2+3)+π΅(β2+2) β1=π΄ Γ 1+π΅ Γ 0 β1=π΄ π¨=βπ Hence we can write (π¦ + 1)/((π¦ + 2) (π¦ + 3) )=(β1)/((π¦ + 2) ) + 2/((π¦ + 3) ) Substituting back π¦=π₯^2 (π₯^2 + 1 )/((π₯^2 + 2) (π₯^2 + 3) ) =(β1)/((π₯^2 + 2) )+2/((π₯^2 + 3) ) Therefore, β«1β(π₯^2 + 1 )/((π₯^2 + 2) (π₯^2 + 3) ) ππ₯=β«1β(β1)/((π₯^2 + 2) ) ππ₯+β«1β2/((π₯^2 + 3) ) ππ₯ =ββ«1β1/((π₯^2 +(β2)^2 ) ) ππ₯+2β«1β1/((π₯^2 +(β3)^2 ) ) ππ₯ By using formula β«1β1/(π₯^2 + π^2 ) ππ₯=1/π γπ‘ππγ^(β1)β‘(π₯/π)+πΆ =(βπ)/βπ γπππγ^(βπ)β‘γπ/βπγ+π/βπ γπππγ^(βπ)β‘γπ/βπγ +πͺ

#### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.