# Ex 7.8, 11 - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by Teachoo

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Ex 7.8, 11 Important You are here

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Last updated at April 16, 2024 by Teachoo

Ex 7.8, 11 ∫_2^3▒𝑑𝑥/(𝑥2 − 1) Step 1 :- Let F(𝑥)=∫1▒𝑑𝑥/(𝑥^2 − 1) We can write integrate as 1/(𝑥^2 − 1)=1/((𝑥 − 1) (𝑥 + 1) ) 1/((𝑥 − 1) (𝑥 + 1) )=A/(𝑥 − 1)+B/( (𝑥 + 1) ) 1/((𝑥 − 1) (𝑥 + 1) )=(A(𝑥 + 1) + B(𝑥 − 1))/(𝑥 − 1)(𝑥 + 1) By Canceling denominator 1=A(𝑥+1)+B(𝑥−1) Putting 𝑥=−1 1=A(−1+1)+B(−1−1) 1 =A ×0+=B(−2) 1=−2B B=(−1)/( 2) Similarly putting 𝑥=1 1=A(1+1)+B(1−1) 1 =A(2)+B×0 1=2A A= 1/2 Therefore, ∫1▒〖1/(𝑥−1)(𝑥+1) =∫1▒〖(1 𝑑𝑥)/2(𝑥−1) +∫1▒〖(−1)/( 2) 1/((𝑥 + 1) ) 𝑑𝑥〗〗〗 =1/2 [∫1▒〖1/((𝑥−1) ) 𝑑𝑥−∫1▒𝑑𝑥/( 𝑥+1)〗] =1/2 [𝑙𝑜𝑔|𝑥−1|−𝑙𝑜𝑔|𝑥+1|] =1/2 𝑙𝑜𝑔|(𝑥 − 1)/(𝑥 + 1)| Hence F(𝑥)=1/2 𝑙𝑜𝑔|(𝑥 − 1)/(𝑥 + 1)| Step 2 :- ∫_2^3▒〖1/(1−𝑥^2 ) 𝑑𝑥=𝐹(3)−𝐹(2) 〗 =1/2 𝑙𝑜𝑔|(3−1)/(3+1)|−1/2 𝑙𝑜𝑔|(2−1)/(2+1)| =1/2 𝑙𝑜𝑔(2/4)−1/2 𝑙𝑜𝑔(1/3) =1/2 𝑙𝑜𝑔[(1/2)/(1/3)] =𝟏/𝟐 𝒍𝒐𝒈 𝟑/𝟐