Examples

Chapter 7 Class 12 Integrals
Serial order wise

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Example 10 Find the following integrals: (ii) β«1β(π₯ + 3)/β(5 β 4π₯ β π₯^2 ) ππ₯ We can write numerator as π₯+3= Aπ/ππ₯ [βπ₯^2β4π₯+5]+ B π₯+3= A[β2π₯β4]+ B π₯+3=β2π΄π₯β4A+B Finding A & B Comparing coefficient of π₯ π₯=β2π΄π₯ 1 =β2A A=(β1)/( 2) Comparing constant term 3=β4A+B 3=β4((β1)/( 2))+B 3=2+B B=3β2=1 Now, we know that π₯+3=A[β2π₯β4]+B π₯+3=(β1)/( 2) [β2π₯β4]+1 Now, our equation is β«1β(π₯+3)/β(5 β 4π₯ + π₯^2 ).ππ₯=β«1β(1/2 [β2π₯ β 4]+1)/β(5 β 4π₯ β π₯^2 ).ππ₯ =β«1β((β1)/2 [β2π₯ β 4]+1)/β(5 β 4π₯ + π₯^2 ) ππ₯+β«1β1/β(5 β 4π₯ β π₯^2 ) ππ₯ =(β1)/2 β«1β(β2π₯ β 4)/β(5 β 4π₯ + π₯^2 ) ππ₯+β«1β1/β(5 β 4π₯ β π₯^2 ) ππ₯ Solving I1 I1 = (β1)/2 β«1βγ(β2π₯ β 4)/β(βπ₯^2 β 4π₯ + 5).ππ₯γ Let t = βπ₯^2 β 4π₯ + 5 Differentiating both sides w.r.t. π₯ β2π₯ β 4 =ππ‘/ππ₯ ππ₯=ππ‘/(β 2π₯ β 4) Now, I1 = (β1)/( 2) β«1βγ(β2π₯ β 4)/β(βπ₯^2 β 4π₯ + 5).ππ₯γ Putting the values of (βπ₯^2β4π₯+5) and ππ₯ I1 = (β1)/( 2) β«1βγ(β2π₯ β 4)/βπ‘.ππ‘/((β2π₯ β4) )γ I1 = (β1)/( 2) β«1βγ1/βπ‘.ππ‘γ I1 = (β1)/( 2) β«1βγ1/(π‘)^(1/2) .ππ‘γ I1 = (β1)/( 2) β«1βγ(π‘)^((β1)/2) ππ‘γ I1 = (β1)/( 2) π‘^((β1)/( 2) + 1)/(((β1)/( 2) + 1) )+πΆ1 I1 = (β1)/( 2) π‘^(1/( 2) )/((1/2) )+πΆ1 I1 = β π‘^(1/( 2) )+πΆ1 I1 = ββπ‘+πΆ1 Putting back π‘=π₯^2β4π₯+5) I1 = ββ(βπ₯^2β4π₯+5)+πΆ1 Solving π°π I2 = β«1βγ1/β(5 β 4π₯ β π₯^2 ).ππ₯γ I2 = β«1βγ1/β(β(π₯^2 + 4π₯ β 5) ) ππ₯γ I2 = β«1βγ1/β(β[π₯^2 + 2(2)(π₯) β 5] ) ππ₯γ I2 = β«1βγ1/β(β[π₯^2 + 2(2)(π₯) + (2)^2 β (2)^2 β 5] ) ππ₯γ I2 = β«1βγ1/β(β[(π₯ + 2)^2 β 4 β 5] ) ππ₯γ I2 = β«1βγ1/β(β[(π₯ + 2)^2 β 9] ) ππ₯γ I2 = β«1βγ1/β(9 β (π₯ + 2)^2 ) ππ₯γ I2 = β«1βγ1/β((3)^2 β (π₯ + 2)^2 ) ππ₯γ It is of form β«1βγ1/β(π^2 β π₯^2 ).ππ₯=γπ ππγ^(β1)β‘γπ₯/π+πΆ2γ γ Replacing π₯ by (π₯+2) and a by 3, we get I2 = sin^(β1)β‘γ((π₯ + 2)/3)+πΆ2γ Now, Putting values of I1 and I2 in (1) β«1βγ(π₯ + 3)/β(5 β 4π₯ β π₯^2 ).ππ₯=(β1)/( 2) β«1β(β 2π₯ β 4)/β(5 β 4π₯ β π₯^2 )γ+β«1β1/β(5 β 4π₯ β π₯^2 ) ππ₯ =ββ(5 β 4π₯ β π₯^2 )+πΆ1+sin^(β1)β‘((π₯ + 2)/3)+πΆ2 =ββ(π β ππ β π^π )+γπππγ^(βπ)β‘γ((π + π)/π)+πͺγ