Find Integration ∫ (x^2 + 1)/ (x^2 + 2) (x^2 + 3) dx

Question 33 Find ∫1▒〖(𝑥^2 + 1)/((𝑥^2 + 2) (𝑥^2 + 3)) 𝑑𝑥〗 Putting 𝒙^𝟐=𝒚 (𝑥^2 + 1 )/((𝑥^2 + 2) (𝑥^2 + 3) )=(𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) ) We can write this in form (𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) )=𝐴/((𝑦 + 2) ) + 𝐵/((𝑦 + 3) ) (𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) )=(𝐴(𝑦 +3) + 𝐵 (𝑦 + 2))/((𝑦 + 2) (𝑦 + 3) ) By cancelling denominator 𝑦+1=𝐴(𝑦 +3) + 𝐵 (𝑦 + 2) Putting y = −3 −3+1=𝐴(−3+3)+𝐵(−3+2) −2=𝐴 × 0+𝐵 × −1 −2=−𝐵 𝑩=𝟐 Putting y = −2 −2+1=𝐴(−2+3)+𝐵(−2+2) −1=𝐴 × 1+𝐵 × 0 −1=𝐴 𝑨=−𝟏 Hence we can write (𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) )=(−1)/((𝑦 + 2) ) + 2/((𝑦 + 3) ) Substituting back 𝑦=𝑥^2 (𝑥^2 + 1 )/((𝑥^2 + 2) (𝑥^2 + 3) ) =(−1)/((𝑥^2 + 2) )+2/((𝑥^2 + 3) ) Therefore, ∫1▒(𝑥^2 + 1 )/((𝑥^2 + 2) (𝑥^2 + 3) ) 𝑑𝑥=∫1▒(−1)/((𝑥^2 + 2) ) 𝑑𝑥+∫1▒2/((𝑥^2 + 3) ) 𝑑𝑥 =−∫1▒1/((𝑥^2 +(√2)^2 ) ) 𝑑𝑥+2∫1▒1/((𝑥^2 +(√3)^2 ) ) 𝑑𝑥 By using formula ∫1▒1/(𝑥^2 + 𝑎^2 ) 𝑑𝑥=1/𝑎 〖𝑡𝑎𝑛〗^(−1)⁡(𝑥/𝑎)+𝐶 =(−𝟏)/√𝟐 〖𝒕𝒂𝒏〗^(−𝟏)⁡〖𝒙/√𝟐〗+𝟐/√𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡〖𝒙/√𝟑〗 +𝑪

Question 33 - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Find ∫ x ² + 1 / (x ² + 2) (x ² + 3) dx

Davneet Singh

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular