CBSE Class 12 Sample Paper for 2023 Boards

Practice Questions CBSE - Maths Class 12 (2023 Boards) →
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Question 31 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by

Find ∫(x 3 + x  + 1)/((x 2 - 1) dx

 

Slide1.JPG

Slide2.JPG
Slide3.JPG
Slide4.JPG
Slide5.JPG

 

Transcript

Question 31 Find ∫1▒〖((𝑥^3 + 𝑥 + 1))/((𝑥^2 − 1)) 𝑑𝑥〗∫1▒〖((𝑥^3 + 𝑥 + 1))/((𝑥^2 − 1)) 𝑑𝑥〗 = ∫1▒〖𝒙+(𝟐𝒙 + 𝟏)/((𝒙^𝟐 − 𝟏)) 𝒅𝒙〗 = ∫1▒〖𝑥 𝑑𝑥〗+∫1▒〖(𝟐𝒙 + 𝟏)/((𝒙^𝟐 − 𝟏)) 𝑑𝑥〗 = 𝑥^2/2+∫1▒〖(𝟐𝒙 + 𝟏)/((𝒙^𝟐 − 𝟏)) 𝑑𝑥〗 = 𝒙^𝟐/𝟐+∫1▒〖(𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏)) 𝒅𝒙〗 Now Solving (𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏) ) = 𝑨/((𝒙 + 𝟏)) + 𝑩/((𝒙 − 𝟏)) (2𝑥 + 1)/((𝑥 + 1)(𝑥 − 1) ) = (𝐴(𝑥 − 1) + 𝐵(𝑥 + 1))/((𝑥 + 1)(𝑥 − 1) ) Cancelling denominator 2𝑥+1=𝐴(𝑥−1)+𝐵(𝑥+1) Putting x = 1 in (2) 2𝑥+1=𝐴(𝑥−1)+𝐵(𝑥+1) 2(1)+1 = 𝐴(1−1) + 𝐵(1+1) 3 = A × 0+2𝐵 3 = 2𝐵 𝑩=𝟑/𝟐 Putting x = −1 in (2) 2𝑥+1=𝐴(=1−1)+𝐵(𝑥+1) 2(−1)+1 = 𝐴(−1−1) + 𝐵(−1+1) −2+1 = A × −2+𝐵 × 0 −1 = −2A 1 = 2A 1/2 = A 𝑨=𝟏/𝟐 Hence we can write it as (𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏) ) = 𝑨/((𝒙 + 𝟏)) + 𝑩/((𝒙 − 𝟏)) (𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏) ) = 𝟏/(𝟐(𝒙 + 𝟏)) + 𝟑/(𝟐(𝒙 − 𝟏)) Therefore , from (1) we get, ∫1▒〖((𝑥^3 + 𝑥 + 1))/((𝑥^2 − 1)) 𝑑𝑥〗 =𝑥^2/2+ ∫1▒(1/(2(𝑥 + 1)) " + " 3/(2(𝑥 − 1))) 𝑑𝑥 =𝑥^2/2+ ∫1▒𝑑𝑥/(2(𝑥 + 1))+∫1▒3𝑑𝑥/(2(𝑥 − 1)) =𝒙^𝟐/𝟐+𝟏/𝟐 ∫1▒〖𝒅𝒙/((𝒙 + 𝟏)) + 𝟑/𝟐〗 ∫1▒𝒅𝒙/((𝒙 − 𝟏)) =𝑥^2/2 +1/2 log⁡|(𝑥+1)|+3/2 log⁡|𝑥−1|+𝐶 =𝑥^2/2+1/2 ( log⁡|(𝑥+1)|+3 log⁡|𝑥−1| )+𝐶 =𝑥^2/2+1/2 ( log⁡|(𝑥+1) (𝑥−1)^3 | )+𝐶

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.