Examples

Chapter 7 Class 12 Integrals
Serial order wise

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Example 16 Find β«1β(π₯^2+ π₯ +1 ππ₯ )/((π₯ + 2) (π₯^2+1) ) We can write equation as (π₯^2+ π₯ + 1)/((π₯ + 1) (π₯ + 2) )=π΄/(π₯ + 2) + (π΅π₯ + πΆ)/(π₯^2+ 1) Cancelling denominator γ π₯γ^2+ π₯+1=π΄(π₯^2+1)+(π΅π₯+πΆ) (π₯+2) Putting x = βπ (β2)^2+(β2)+1=π΄((β2)^2+1)+0 4β2+1= 5A 3/5 = A Putting x = π π₯^2+ π₯+1=π΄(π₯^2+1)+(π΅π₯+πΆ) (π₯+2) 0+0+0= A(0 + 1) + (0 + C) (0 + 2) 1 = A + 2C 1 = 3/5 + 2C 1 β 3/5 = 2C 2/5 = 2C C = 1/5 Putting x = 1 π₯^2+ π₯+1=π΄(π₯^2+1)+(π΅π₯+πΆ) (π₯+2) 1+1+1= 2A + (B + C)(3) 3 = 2A + 3 (B + C) 3 = 2(3/5) + 3 (B+1/5) 3 β 6/5 = 3 (B+1/5) 9/5 = 3 (B+1/5) 3/5 β 1/5 = B B = 2/5 Thus, (π₯^2+ π₯ + 1)/((π₯ + 1) (π₯ + 2) )=π΄/(π₯ + 2) + (π΅π₯ + πΆ)/(π₯^2+ 1) (π₯^2+ π₯ + 1)/((π₯ + 1)(π₯^2+ 1)) = 3/(5 (π₯ + 2)) + (1 (2π₯ + 1))/(5 (π₯^2 + 1)) Hence, our equation becomes β«1β(π₯^2+ π₯ + 1)/((π₯ + 2) (π₯^2 + 1)) ππ₯= β«1β3/(5(π₯^2 + 1)) ππ₯+β«1β1/5 ((2π₯ + 1))/(π₯^2 + 1) ππ₯ = β«1β3/(5(π₯^2 + 1)) ππ₯+ 1/5 β«1βγ2π₯/(π₯^2 + 1) ππ₯+γ 1/5 β«1β1/(π₯^2 + 1) ππ₯ ππ 1/5 β«1β2π₯/(π₯^2+ 1) ππ₯ Let π‘=π₯^2+ 1 ππ‘/ππ₯=2π₯ ππ‘=2π₯ ππ₯ Substituting, =1/5 β«1βππ‘/π‘ = 1/5 log |π‘| + C_2 = 1/5 log |π₯^2+1| + C_2 ππ 1/5 β«1β1/(π₯^2+ 1) ππ₯ = 1/5 γπ‘ππγ^(β1) (π₯)+C_3 Hence β«1β(π₯^2+ π₯ + 1)/((π₯ + 2) (π₯^2+ 1)) ππ₯ =π/π πππ|π+π|+π/π πππ|π^π+π|+π/π γπππγ^(βπ) (π)+ C where C = C_1+ C_2+C_3