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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.