Ex 7.5

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.5, 9 Integrate the function (3π₯ + 5)/(π₯^3 β π₯^2 β π₯ + 1) Let I=β«1β(3π₯ + 5)/(π₯^3 β π₯^2 β π₯ + 1) ππ₯ We can write integrand as (3π₯ + 5)/(π₯^3 β π₯^2 β π₯ + 1)=(3π₯ + 5)/(π₯ β 1)(π₯^2 β 1) =(3π₯ + 5)/(π₯ β 1)(π₯^2 β 1^2 ) =(3π₯ + 5)/((π₯ β 1) (π₯ β 1) (π₯ + 1) ) =(3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 Rough π₯^3βπ₯^2βπ₯+1 Put π₯=1 1^3β1^2β1+1 =1β1β1+1 =0 So, π₯β1 is a factor of π₯^3βπ₯^2βπ₯+1 We can write it as (3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 =π΄/((π₯ + 1) ) + π΅/((π₯ β 1) ) + πΆ/(π₯ β 1)^2 (3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 =(π΄(π₯ β 1)^2 + π΅(π₯ + 1)(π₯ β 1) + πΆ(π₯ + 1))/(π₯ + 1)(π₯ β 1)(π₯ β 1) (3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 =(π΄(π₯ β 1)^2 + π΅(π₯^2 β 1) + πΆ(π₯ + 1))/((π₯ + 1) (π₯ β 1)^2 ) By Cancelling denominator 3π₯+5=π΄(π₯β1)^2+π΅(π₯^2β1)+πΆ(π₯+1) Put π₯=1 in (1) 3Γ1+5=π΄(1β1)^2+π΅(1^2β1)+πΆ(1+1) 8=π΄Γ0+ π΅Γ0+πΆΓ2 β¦(1) 8=2πΆ πΆ=4 Putting π₯=β1 in (1) 3π₯+5=π΄(π₯β1)^2+π΅(π₯^2β1)+πΆ(π₯+1) 3(β1)+5=π΄(β1β1)^2+π΅((β1)^2β1)+πΆ(β1+1) β3+5=π΄(β2)^2+π΅(1β1)+πΆ(0) 2=4π΄+π΅Γ0+πΆΓ0 2=4π΄ π΄=1/2 Putting x = 0 in (1) 3π₯+5=π΄(π₯β1)^2+π΅(π₯^2β1)+πΆ(π₯+1) 3(0)+5=π΄(0β1)^2+π΅(0β1)+πΆ(0+1) 5=π΄βπ΅+πΆ 5= 1/2 βπ΅+4 5= βπ΅+9/2 π΅=9/2 β5 π΅=(β1)/2 Hence, we can our equation as write (3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 =π΄/((π₯ + 1) ) + π΅/((π₯ β 1) ) + πΆ/(π₯ β 1)^2 (3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 =((1/2))/((π₯ + 1) ) + (β 1/2)/((π₯ β 1) ) + 4/(π₯ β 1)^2 (3π₯ + 5)/γ(π₯ + 1) (π₯ β 1)γ^2 =1/2(π₯ + 1) β 1/2(π₯ β 1) + 4/(π₯ β 1)^2 Integrating π€.π.π‘.π₯ I=β«1β(3π₯ + 5)/(π₯^3 β π₯^2 β π₯ + 1) ππ₯ =β«1β(1/2(π₯ + 1) β 1/2(π₯ β 1) + 4/(π₯ β 1)^2 ) ππ₯ =1/2 β«1βππ₯/(π₯ + 1) β 1/2 β«1βππ₯/((π₯ β 1) )+4β«1βππ₯/(π₯ β 1)^2 Hence I=I1βI2+I3 Now, I1=1/2 β«1β1/(π₯ + 1) ππ₯ =1/2 logβ‘|π₯+1|+πΆ1 Also, I2 =1/2 β«1β1/(π₯ β 1) ππ₯ = 1/2 logβ‘|π₯β1|+πΆ2 And, I3=β«1β4/(π₯ β 1)^2 ππ₯ =4β«1β1/(π₯ β 1)^2 ππ₯ =4β«1β(π₯ β 1)^(β2) ππ₯ =(4(π₯ β 1)^(β2 + 1))/(β2 + 1) +πΆ3 =(4(π₯ β 1)^(β1))/(β1) +πΆ3 =(β 4)/(π₯ β 1)+πΆ3 Therefore I=I1βI2+I3 I=1/2 logβ‘|π₯+1|+πΆ1β 1/2 logβ‘|π₯β1|βπΆ2+(β 4)/(π₯ β 1)+πΆ3 =1/2 logβ‘|π₯+1|+β 1/2 logβ‘|π₯β1|β4/(π₯ β 1) +πΆ1βπΆ2+πΆ3 =1/2 [logβ‘|π₯+1|βlogβ‘|π₯β1| ]β4/(π₯ β 1) +πΆ =π/π πππβ‘|(π + π)/(π β π)|β π/(π β π) +πͺ