Ex 7.8

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.8, 11 âŤ_2^3âđđĽ/(đĽ2 â 1) Step 1 :- Let F(đĽ)=âŤ1âđđĽ/(đĽ^2 â 1) We can write integrate as 1/(đĽ^2 â 1)=1/((đĽ â 1) (đĽ + 1) ) 1/((đĽ â 1) (đĽ + 1) )=A/(đĽ â 1)+B/( (đĽ + 1) ) 1/((đĽ â 1) (đĽ + 1) )=(A(đĽ + 1) + B(đĽ â 1))/(đĽ â 1)(đĽ + 1) By Canceling denominator 1=A(đĽ+1)+B(đĽâ1) Putting đĽ=â1 1=A(â1+1)+B(â1â1) 1 =A Ă0+=B(â2) 1=â2B B=(â1)/( 2) Similarly putting đĽ=1 1=A(1+1)+B(1â1) 1 =A(2)+BĂ0 1=2A A= 1/2 Therefore, âŤ1âă1/(đĽâ1)(đĽ+1) =âŤ1âă(1 đđĽ)/2(đĽâ1) +âŤ1âă(â1)/( 2) 1/((đĽ + 1) ) đđĽăăă =1/2 [âŤ1âă1/((đĽâ1) ) đđĽââŤ1âđđĽ/( đĽ+1)ă] =1/2 [đđđ|đĽâ1|âđđđ|đĽ+1|] =1/2 đđđ|(đĽ â 1)/(đĽ + 1)| Hence F(đĽ)=1/2 đđđ|(đĽ â 1)/(đĽ + 1)| Step 2 :- âŤ_2^3âă1/(1âđĽ^2 ) đđĽ=đš(3)âđš(2) ă =1/2 đđđ|(3â1)/(3+1)|â1/2 đđđ|(2â1)/(2+1)| =1/2 đđđ(2/4)â1/2 đđđ(1/3) =1/2 đđđ[(1/2)/(1/3)] =đ/đ đđđ đ/đ